Integrals POINTS TO 1. Antiderivative (or Primitive): A function (2) is said to be antiderivative or primitive of a function fe)i€ve) =fadie, A191) = fe). 2 For example, 7 is primitive or antiderivative of x because Similarly, F a(x ) if . Similarly, d(Frc syatoex In this way, we see that a function has infinitely many antiderivatives or primitives. ie, if $(x) be an antiderivative of fix), then 6(x) + C is also antiderivative of f(x), where C is any constant. Because, Live+ci=o@+0=9@ =f@) Indefinite Integrals: If f(x) is a function, then the family of all its antiderivatives is called Indefinite Integral of fix) It is represented by J f(x)dx (read as indefinite integral of f(x) with respect to 2), 4 For example, Pax = oe G fede Z+c “\l 2 z= | n ool ©) A ful as ial < 16] m ve) ) Why is it called Indefinite Integral? It is called indefinite because it is not unique. Actually there exist infinitely many integrals of each function, which can be obtained by choosing C arbitrarily from the set of real murmbers, 2. Some Standard Integrals: 0 frtar= ZL ecoey @ fH =t0g |x [+C sr = log ii) fdx=xtC (iv) fcosxdx = sinx+C (@) Jsinxdx =-cosx+C (wi) [sec?x dx = tanx+Cee ir [aa] 2 Ww 5 rT (at e) = va) lal = ©) ja (vif) Jcosec?x dx =-cotx+C (viii) [secxtanx de = secx+C (ix) Scosecxcotx dx=-cosecx+C (x) fetdx =e" +C i) fatde=— +0 Toga aid J Epes sin tne (i) @) J pes tanta - zy eu) | 3. Methods of Integration: It is not possible to integrate each integral with the help of following methods but a large number of various problems can be solved by these methods. So, we have the following methods of integration: (i) Integration by Substitution. (i) Integration by Parts. (iii) Integration of Rational Algebraic Functions by Using Partial Fractions. 4. Integration by Substitution: The method of evaluating integrals of a function by suitable substitution is called Integration by substitution. We therefore give some of the fundamental integrals when x is replaced by ax + b. ea o Jes tae OO sn tt i) Seger = Fos | @ + [+c Gi) fertae=hertac (io) Jade => re +Ca> anda #1 (@) Jsin(ax+ bydx =—Leos(ax+ b+ (oi) feos(ax+ dx = bsin(axt hy +c (oid) fsecax+ Bde = Htan(ax+h)+C (will) fecosee*(ax+ Bide = ~f cot ax +8) +C (a) Jscc(ax + bytan(ax+ bydx=1sec(ax +8) +¢ (2) foosec(ax+ #) cot(ar + 8)dx = cosec(ar+h) +C Ge) ftan(ax+8)dv=—Hog | cos(ax+ iy |+C= 1 toglsec(ax +6) |+C (ai) feot(ax + b)dx = log | sin(ax +b) |+C 5. More Standard Results: Jtanx dx =-log | cosx |+ C= log | secx |+C, provided x is not an odd multiple of = Jootx dx = log | sin |+C(GBke +c Jscex x= loglseex +tanxl+C= log Jcosec x dx = log | cosec x - cotx |+C= logan 6. Integration by Parts: To integrate the product of two functions, we use integration by parts. The method is as given below: Let u and v be two functions of x then Juvde=ufode— I{ Gt Joaspar Note: ( To integrate the product of tvo functions we choose the 1st function according to word ILATE, where I stands for inverse function, L stands for logarithmic function, A stands for the algebraic functions, T stands for trigonometrical function arid E stands for exponential function. (i sf the integrand has only one function ther unity, ie, 1 i taken tobe the second function. (Gi) Integration by parts isnot applicable to product of functions in all cases. For example, the method does not work for [.sinx dx. The reason is that there does not exist any function whose derieative is Asin. (G0) Observe tha while finding the integral ofthe second function, we do not add a constant of integration on both the sides. 7. Results of Some Special Integrals: u = rus I Va) o (e) z2) m a m K< w m ys) © Io a = log|S24|+c +#C or log | x+ Ve? +a? |+C gf | Cortog | xt V2 =a? |+C 2 () [We Pde = Fa? x? +S sin (i) [VP ede = 5 (in [Va ar =F Eft aH + Dig | x+ aoa Itc Theorem 1. The indefinite integral ofan algebraic sum of two or more functions is equal to the algebraic sum of their integrals, ie, Jf) + Olde = J foeyde+ | geyde Theorem 2. A constant term may be taken outside from the integral sign ie, if k isa constant then Jk fe)dx = kf fx)dx Jog | Poa [+ca Ww co Py ee 2 Wu jad (e) ps WY) lal = (e) oO Theorem 3. if the numerator in an integral is the exact derivative of denominator, then its integral is logarithamic of denominator, . Fadi _ ie, J fay 2B 1Aa) IFC Theorem 4, To integrate a function whose numerator is unity and denominator is a homogeneous function of degree in cos x and sin x ie, the integrals of these forms: fi dx f dx f dx dx dx a+bsinx” acosx+bsinx” asinx + beosx a+ beosx” asiny +b" i.e, when integrand is a rational function of sin.x and cos x. To find these, we can use following substitution. (0 By putting a = r cos a, b = r sin a respectively according to question OR ond (ud) 5 (ii) By putting sinx = —— >, cosy= >and putting tan3;= and then 1+tan?> '1+tan? >) simplify. 2 ( 2) Theorem 5. To integrate a function whose numerator is 1 and denominator isa homogeneous function of the sevond degree in cos x and sin x or both, ie., To evaluate such type of integrals we proceed as follows: (@ Divide the numerator and denominator by cos” x and then (ii) Putting tan x=z or cotx = zand then simplify. Theorem 6. Integrals of the type |e“? ¢” (xydx,| f’ (x)cos[f(x)]dx,/ sin [f()]f" dx, Slogtfaylf' (pax. To evaluate these type of integrals, put flx) = t so that f'(x) dx = dt and then integral converts to the standard forms for which the integrats are known. Note:If the integrand is a rational function of ¢, then it always needs a replacement as the differentiation and integration of e*is the same. Thus, if on substituting denominator = ¢, the derivative of denominator is not present in the problem, then we need to generate it by multiplying and dividing by a suitable term containing the exponential function in numerator and denominator. »b Integration by Partial Fractions 8. Rational Function: Rational function is defined as the ratio of two polynomials in the form of me where P(x) and Q(x) are polynomials in x. If the degree of P(x) is less than degree of Q(x) then it is said to be Proper, otherwise it is called an Improper Rational Function. Thus if ae is improper, then by long division method it can be reduced to proper function ie, Pe) ‘ 7 Ae) Quy 77 ie 7 — Tea) is a function of xand-Ge fractions can be evaluated by breaking in factors given as follows: is a proper rational function. Such||S.No. | Form of the rational function __Form of the partial fraction. px tg G-ae-y tt? 1 petg (=a? mitqetr petgrtr A, (—a)°G—5) =a) @—a? 9) px tga tr A i ye (e-a°@-9) €-9 Ga Ga & pesgrtr A_, Brtc (me thet) &- D) where x? + bx +c cannot be factored further| torte! The constants A, B, C, etc. are obtained by equating coefficient of like terms from both sides or by substituting any value for x on both sides. To find the integral of the form { we write te ax?+bxte u = rus I Va) o (e) z2) m a m K< w m ys) Now putting x+:2-= so that di = dt, Therefore, writing {~-2=k, and find the integral of dt reduced form . Tel’ Pew pet ). Integrals of the form: Ime Step I. The numerator px + q is written in the form pet qe AL ax hxc) +B > prtq= A(Qax+b)+B Step Il. The value of A and B is obtained by equating the coefficients in the above equation. Step ILL (px + q) is replaced by A@ax + 6) + B and we write the given integral as, + A(ax+b)+B f ia 4) wef ests a ax? bx te ax? thx +e (ets . Tntegenls nf the fons. | ASE dx. eet prt q= Abe shee +B = px+q=AQax+b)+B Step II. The values of A and B are obtained by equating the coeffcients in the above equation. Step IIL (px +4) is replaced by A (2ax + b) + Bin given integration as (px+9) AQax+b)+B SaaS Se Vax? bxte Vax?+bxee ‘dx and then solved.ee ir [aa] 2 Ww 5 rT (at e) = va) lal = ©) ja 9 11. Integration of the form let where p(x) and q(x) are polynomials such that degree of p(x) 2 degree q(x). Step I. p(x) is divided by q(x) and it is written as px) Ra) Way 22) Gay Where QC) is quotiont polynomial and RG s remainder polynomial R(x) ax) P(x) P(x) Step II. 7@* replaced by (a+ 2) 2 as (AS 12. Integral of the form {sin™x.cos"x dx () Ifthe exponent of sin x is an odd positive integer, then put cos x =f. (i) If the exponent of cos xis an odd integer, then put sin x = f. 13, Integral of the form { e*(f(x) +f’ ())de = fia).e*+ C ) Definite Integrals 14. Definition: If F(x) is the integral of f(x) over the interval (a, b), ie, { f(x)dx = F(x) then the definite integral of fx) over the intereal [a,b] is denoted by f fla) is defined as dx = (ow Jae and then solved. | eye Fe) -Fle where ‘a’ is called the lower limit and 'b’ is called the upper limit of integration and the interval (a, b) is called the interval of integrations. 15. Some Useful Results: The following results will be useful in evaluating the definite integrals as the limit of sum. n(n-1) () Y@r-1) = 142434..4 01-1) = (2!) cosa+ cos(a+h)+ cos(a+ 2h) +... + costa + (n—1)h} = 16. Fundamental Properties of Definite Integrals: There are certain properties of definite integrals ‘which can be used while solving the definite integral5 : @ J fedare J fleyie (Change of variable) (i J pxyaee— | playa (inter change the limits) 7 > Gif fla)ae=f flxyde+ f fxyde, whereace dx, x#2is equal to [CBSE 2023 (65/5/1)] @1 () -1 (2 @2 ax . f Gin ca)sin@ =) i eaual to INCERT Exemplar] ine alogl 22-9 a.c 5 b-aylog| 8 é @ sinb-a)log| Ga | © cosecte—a)log| Gy | % b-atog| Lec i sintb—aytogl = |. © cose a)log| Gay sino a)tog| 5 Stan” Vxdx is equal to INCERT Exemplar] (@) (e+ Itant vx -y/x+C (b) xtan* Vx-Vx+C (©) ¥x—xtant ¥x+C d) vx-(e+ Iter Vr +C ae fet ( be = dx is equal to INCERT Exemplar] e Ware or ©. @ Gate [2*"*ax is equal to [CBSE 2023 (65/3/2)] oe ” 22, a o @ 24C (b) 2*7log2+C (0 gate @ 2aeete J sec*x dx is equal to [CBSE 2020 (65/1/1)] a (@ -1 mo ol @2 The integral dx is equal to [NCERT Exemplar] ap @ Herd)’ © Hard) 'ec ry arte @ L{Ses) +e ri tan’ (2x) dy is equal to [CBSE 2020 (63/4/1)] a @ = @ © a = JES foddx is equal to INCERT Exemplar] @ Pfe-od — ® Pfarodx E fara @ IE fear If fand g are continuous functions in [0, 1] satisfying f (x) = fla —x) and g(x) + g(a-x) = a, then { fe).g()dx is equal to INCERT Exemplar] > ©) Sh fedax (© § fear @) aff fox)de a x elxle1 2 G Fun is equal to INCERT Exemplar] (@) log2 (b) 2log2 (©) Fiog2 @ 4log2ae 17. Gg Spat avthen Ede is equal to [NCERT Exemplar] 2 e e (@ aie to ari-$ @at-$ @ att+$ 18, |xcosxx|dx is equal to INCERT Exemplar] 2 8 4 2 1 @ & wt @2 @t 1. upad= x44, then fixlis ICBSE Sample Paper 20231 @ Petogltl+C — Frtogixl+c ©) F+toglel+C — @ F-loghxl+c 20, The value of eat [CBSE Sample Paper 2023] 3 1 9 @) log © log 5 (@ Flog? @ tog 2 a. | /4-x7dx equals (CBSE 2023 (65/3/2)] 3 (@) 2t0g2 () -2l0g2 oF @n 22. 1f Zea) =10g x, then fad equals: ICBSE 2023 (65/1/10) w -hec @xoge-D+C — @allogren+e @ 4+¢ a. f “ee 2 jae is equal to: [CBSE 2023 (65/1/2)1 @ - O -z () v3 @ -v3 24. If at [flx)] = ax +b and (0) =0, then f(x) is equal to [CBSE 2023 (65/2/1)] 2 was ) ste © Esmee we 25, Antisderivative of @"=—~ with respect tox is {CBSE 2023 (65/2/1)1 @) sece(F-x}+c ( ~se2(F =x) +¢ (© togsee($-x)]+ (@) ~og|see(4-x}]+c SCX dx equal: CBSE 2023 (65/51: 26, [ate dr equals l (65/5/01 @) seex-tanx+C (6) seex+tanx+C — (e) tanx-seex+C (@) (seex+tanxy+C Answers 1@ 2 (@) a@ 40 5 @ 6.) 7.0 8. @) 9. @) 10. (c) 1. @ 2. (i) 13. (@) 14. @) 15. () 16. () 17. (b) 18. @) 19. () 20. (¢) 21. (i) 2.) 23. (@) 24. (6) 25. (0) 26. (b)Solutions of Selected Multiple Choice Questions 4. Wehave, I= xe" dx st = 3xdx=dt aeaet a elas ogee feldt=teec Option (a) is correct. eats 08"(xe") meet a (xe +e%)de = dt = (eter = dt 2 Let l= a dx eect = [= Jsec%t dt=tant+C= tan (xe!) +C Orton (@)is correct 2a + cos? sin2x + costs) dx a Rie eee Sintxcostx / sin?xcos?x = [sec?xdx + { cosec*xdx = tanx -cotx+C ©. Option (c) is correct. #2 x-2 ifr-220 = (x=2) ifx-2<0 le, Ile 272 BOR Bey (EON -e-2) ifr <2 1 ee 3 ope [Die fea 4 Ds =-[1-Cn]=-2 * Option nace H Teal sin(b—a) dx sing — Se 5 “Gabe a)’ sin(&— a)sin(x—6) 1 sin(x-a—x+b) sin(b—a)’ sin —a)sin(x —b) = __1_Sin(a =a) costa - 8) costa) sin (xB) ~ sin(b=a) sin (x —a) sin (x —b) ‘Slcot (x -b) —cot (x-a)]dx = tt * sin(b—a)7 licg| sine—b)] log sin(e= a) []+C = cosec( —a) log] ste“ Option (c) is correct. 10. f2t?dx=f2*.2dx=2?[2%dx a Be 2? Pxag triage te Option (c) is correct. 1. Wehave | sec*x dr =[tan x], i = tan} —tan(—T)=141-2 *. Option (a) is correct xs xs 13. J tan®(2x)dx = J (sec?(2x)-1)dx é é fs ef -. Option (a) is correct. a. 12 J fongcur = fpa-agie—node= | fora gtaper sal porde-f forgery > raf foydx-1 > 1=-4f ponds i ! 1 +. Option (6) is correct. Palzle1 te p= fH, ax? +2[x/+1 2 Isle ele sanded "4 Baader 87s Gey lodd function + even fisnction] ep Ft 5 sh Be, 3 1 on] 2h ten a) te =2loghx + 1]|, =210g2. ©. Option (b) is correct. a 2 18. Since T= [|xcosnx|dx = 2 {|xcosnxldx se ° = 2} f bxcosmelar+ f ixcosnr|ax+ [|xcosmelax 2 2 *. Option (a) is correct.F@paxet, weget Integrating both the sides wart. x, we get Spee s(r+d a = Fl) +Hogizl +c <. Option (6) is correct. dx 13 2% 1 s 1 1, (10 FEEL wed fogs? +f -£ log10—10g5) Log 2) 1 log? :. Option (c) is correct, We have, 2 a f ¥4-2dx= f QP =x e Ea 77% =2sin'1=2x On integrating both sides, we have = fix) = flog xi > fo=logx.fidr-j(H togs.fudr)tr = Adtlogexe—fLaede oe flx)= xlogx— fdx=xlogx—x+C = ft)=x(logx-1)+C -. Option (b) is correct. GP see(x-F oo(f-f)-ml Given, A lp|=ar+band f0)=0 ola On integrating both sides, we have flxd= [laxtbde= "> +be+C > f= + ec (i) ‘Also, f(0) = 0, we have from (i) A= Os px04e >0=C Putting in (i), we have 2 jaye tbe ++ Option (b) is correct.25, We have, fanx-1 Jaane = log, +c x see(—x) * Option (¢) is correct. sec sec x(soc y+ tan x) ae la sec? x—tan? x J(sec?x+ sec x tan x)dx sect de [sec xtan dx =tanx+secx+C 2 Option (@)iscorrect Assertion-Reason Questions The following questions consist of two statements—Assertion(A) and Reason(R). Answer these questions selecting the appropriate option given below: @ Both A and R are true and R is the correct explanation for A. © Both A and R are true but R is not the correct explanation for A. © Ais true but Ris false. @) Ais false but R is true. 8 yi0-x Assertion (A): J e+ fae" * 5 Reason (R): | flvdde= J flatb—xdx [CBSE 2023 (65/5/01 2. Assertian(A) : I six = tantx+C, where C is an arbitrary constant. ¥ 1 lex! dx = tan*x+C a Reason (R) : Since 7tan’ Ta 3. Assertion(a) : Jdx= z+ vx Reason (R): fcosxdx=sinx+C + tan. 4, Assertion(A) : If f'(x)=x+ yagi and f(O)=Othen fla) = a wt Reason): fades 2 +C 5 Assertion(a) +f cosedx=1 6 Reason (R) : lfflz)is continuous in [a,b] and { flx)dx= (2), then | fla)de= 6(0)- 4).6. Assertion(a): 1 Reason (R) : Bhs Tree } @ v1-x X46 rp 7. Assertion(A): J /1+cosx dx=2 Reason (R) : 8 Assertion(A) + = loglr+ yx? +a? |+C 1 3 (log xtc Reason (R) : flare logx+¢ Answers 1 @ 2 (@ 3. (b) 40) 5 (a) 6.0) 7.0) 8. (b) Solutions of Assertion-Reason Questions 3 dx (i) V10~(8+2—x) 2 Wer2-x+V10-@+2-2)™ 2 VlO—x+ Vx a Adding (i) and (i), we get fyvx+vi0 abit = eros [deol =8-2=6 = 1=3 So statement A is correct also statement X is true and gives correct explanation of the statement A. . Option (a) is correct. Clearly, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A) ++ Option (a) is correct. =fa(x) =v¥x+C Wehave, > 2x Clearly, both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A). ++ Option (b) is correct. Wehave, f'(x)=x+ 1+x? aw fo =F eoae=f{e+( 3 fetter txt 0+tan"0+C=04+C=C 2 = fay=Fttante+c - fO=0 > FO)2 3 02C 2 fatten ts Clearly, both Assertion (A) and Reason (R) are true but Reason (R) is not correct explanation of Assertion (A). +. Option (6) is correct. 5. Wehave, | cosx dx =[sin xh’? = sin ~sin0=1-021 é Clearly, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). ++ Option (a) is correct. a 6. Wehave, | 0 y1-2 ‘ax =[sin“1x], = sin'™1 - sin 70 Clearly, both Assertion (A) and Reason (R) ate correct and Reason (R) does not gives correct explanation of Assertion (A). <. Option (b) is correct. zp) re my 7 J] areosx dea] 2008 (F)ar cos} dr 4/2 [sin x54 al =22| a 4| 2 So statement A is correct. Also statement R is correct but R does not gives correct explanation of A. * Option (6) is correct. (log x)" a reso = te 1 Putlog x=t stdr=dt apeta ce (oe). Pate 5 ecee Sostatement A is correct. Allso statement R is correct but R does not give explanation of statement A. -. Option (b) is correct. CONCEPTUAL QUESTIONS 1. Evatuate: ;— [CBSE 2020 (65/3/1)) 9+ ax de y Sol. f sae als tan’ Wen [CBSE Marking Scheme 2020 (65/3/1)]atttiget 2 Find: | de [CBSE 2020 (65/4/1)] oe 1 [CBSE Marking Scheme 2020 (65/4/1)] Sol. I= s(269- F@))ar=- af se INCERT Exemplar] 42 Sol. Let ria 3 ai(e-1+ Sy )ae + 2 =Je-Dde+3 (de =A xt Bogle 11+C [CBSE 2020 (65/111) wid __ [Toppers Answer (65/1/1) 2020) (CBSE 2020 (65/1/1)] [Topper’s Answer (65/111) 2020][CBSE 2020 (65/1/1)] [Topper’s Answer (65/1/1) 2020] [CBSE 2020 (65/4/1)] % % [CBSE Marking Scheme 2020 (65/4/1)] 8 Find the value of {"|x-5 | dx [CBSE 2020 (65/5/0)1 + 7 15 Sol. {lx—5ldx = { 6-x)dx => Mth i 1 [CBSE Marking Scheme 2020 (65/5/1)] 9. Find [x*logx dx [CBSE 2020 (65/1/1)] w w [CBSE Marking Scheme 2020 (65/1/1)] Very Short Answer Questions [CBSE 2021-22 (Term-2)] = wherer-2=f >dr=dt ¥QP-(x-2F wa (f)-cnn({CBSE 2021-22 (Term-2)] + [a ac _ . Cwnee Cte canton” 8] er [Topper's Answer 2022] Evaluate: { x*e*dx (CBSE 2021-22 (Term-2)] a 1 1 Sol. J re%de=[x7e%],- J 2xe"dx 1 ° a xe" dee" + 207] % of % [CBSE Marking Scheme 2021-22 (Term-2)] [CBSE 2020 (65/2/01 Sol. Given Integral is % Let cos x= = sin dr == dt % % +c w ICBSE Marking Scheme 2020 (65/2/1)]x Find J ds fPtaee2 ee Ln (ae Sel. Terese Tenara® Ne x+2 =-log [x+1| +2log [x+2|+C ax [CBSE 2020 (65/5/1)1 1 1 ICBSE Marking Scheme 2020 (6515/1)] Sol. 7. Find the value of [ x(1-x)"dx. é so. fxa-ntér=fa-na-1+ntdr= {etx Ndr ao ng eo a Fl n+l m2 ntl at2 (tiynt2) [CBSE 2020 (65/5/1)) % % wim ICBSE Marking Scheme 2020 (6515/1)] [CBSE 2020 (65/5/1)1 wan ICBSE Marking Scheme 2020 (6515/1)] (CBSE 2019 (6515/1)) [CBSE 2019 (6515/1)1 & Find (oe z=i eel Bek Ee ea, _ seat -1, 2 (x-2)(x-3) (x2) (3) = yt al pity == log (e-2) + 210g -3)+C ek tee top te -3)2+C= & [a6 * Jog (x - 2) +]og(x -3)?+C 9%. ox so. fin ents) 7) x = toge(+3)],° =e 2) -1ogse( 4-2 = log(s/2) log (s2e(0)) = log(v2)—logt 1 = logy log 2 x )40. Find: /——“*_ [CBSE 2019 (65/4/1)) V5~4x-2 dx dx sol [—— - | y5-4x-2x? © /7-2-4r-207 dx y7-2(14 2xtx) Zeen)sc v1+sin2x Line feo ICBSE Sample Paper 2018] Viens Sol I=Je' oy & lg? tco jx Wisin? x + cos? x +2sin x. cose “ie T+c08 2 ee erg Vinx +c0sz)? | sinx+cosx ae aoe lO ots @ Sot )ae cos? =} Jet (see see. tan x) de =p eseerte belega+rend 12. Evaluate: Fae [CBSE Dethi 2012] % Sol. Let t=tan"x = dt=—1 Sax tee Also when, x=0,1=Oand when x=1, t= atantx f Tees tat -Ef" Ue x 2h “ale °}" 32 13, Evaluate: (' ICBSE (F) 20091 V0xt3 x Vax 43) Vdx 2x3 a+ 3247 farsa? . a RPS) a 5g 5/3 (-4+2)« zr1 14, The acceleration of an object is given by a(t) = cos(xt), and its velocity at time t= 0 is E-. Find the net distance travelled in the first 1.5 seconds. Sol. We know that, velocity is given by 0) = 900) fatudte where xt) is velocity function, 2(0) is initial velocity and a(x) is acceleration. v(t) = 37 f costa 1 fsinm} 1,1. x BS [-aet rainy *. Net distance travelled is 3 Pads = 5(1.5)-8()=s(5)-s@=f (4+sinatat (1.5)-8@=3(3)-se= f [3+sinm) if 2 eee 15 ‘e*(cosx~ sina) cosec*x dx [CBSE 2019 (65/5/1)] [NCERT] Sol. Let I= [e*(cosx—sinx) cosec*x dx = { e*(cotx. cosecx ~ cosec x) dx = Se*coseex cotx dx — | ¢*cosecx dx mi (er Rescrees meee" -c3. | emer [Using integration by parts for 2nd integral] Je cosecx.cotx dx — e"cosecx + C- Je" cosecx.cotx dr ~ecosec x +C. inde (£082¥ = €0820 16. Find: |e ede (CBSE (AI) 2013] Sol Let f= [£08282 20828 py _ j Qeos'x-1)-Qeostart) ,,_ 4p c0s"x-cos'a cosx— cosa cosx—cosa cosx— cosa = pp (eos —cosa)(cosx+eosa) =2f Pa dx =2{(cosx + cosa)ax = 2fcosxdx+2cosa J1dx =2sinx+ 2x cosa+C Short Answer Questions 1. Find: [e*.sin2xdx {CBSE 2021-22 (Term-2)] Sol. Let 1=Je*.sin2xdx 10 Using ILATE we have Ta sin2x feta —j( TR. ferax)te a I= sin 2x e* — [2cos2x e*dx = sin2xe* —2 fcos2x ede deos2x dx = Tesin2xe"—2]cos2e fej Jef dx)ax] (Again using TLATE)* sin 2x - 2[cos2x.e* -[—2sin2x.e“dx] T= e*sin2x - 2e*cos2x- 4/sin2x “dx *(sin2x - 2cos2x)—41+C, 5I= e*(sin2x - 2cos2x)+C, S. 5 5 (sin2x — 2eos2x)+ 5 (sin2x ~ 2cos2x)+C 2 Find: J [CBSE 2021-22 (Term-2)] 2 Gn @ea) Sol. Wehave,I={ aie dx Letx?=t => Grits 2 1 “hea Swann pop lt2)=04 aera # “paral lala = toilet bela 2 ett =log| $3 |+C=10g ies 2x Paes as zea |"C @ 3. Find: |g [CBSE 2023 (65/5/1)] gf Sok I= \ ogee a oi P-PeeDyY eet @-DGF+1) Pow 4x-1 Boe oe 1 1 Boe = x4145 = rt eaeee Pov ex1 GDF +1) x Pa ee lax = fede +fae+ a 1 r+1+—__] —— {aya een ® seth 1 ey 4 A Brte (e102 41) F-1 7 Pa AG?+1)+(Br+C)x-1) _ Ax?+A+Bx?+Cx-Br-C (Di? #1) @ Yx?+1) Where h = J_ (A+B)x*+(C-B)x+(A-O) 7 = DG2+1) = 1s (A+B)? + (C-Bx+(A-O) Equating both sides like powers of x, we get oy yef ara] + Lanct = Floghe -1|- Floglx?+1|+ Ftan"tx +c 2 x 1 1 14 S4x+Htoghe- 11 Floghy? + 11+ F tant +C 2 . 1 4 Fin JaWe+DWz*D dx [CBSE 2023 (65/1/1)) 4. Sol. Wehave,| Te aayygeny vet Vest actede=dt Lede = aut Be ve 1 “VGnesn* wo (tt2)—(ttY) fd 1 Gey tt Sa) = aflog|#+1]-logle+2l]+C t+1 yrt+1 nat {—___* __ & Find: J yaa) caste=b) dx [CBSE 2023 (65/1/1)] 1 Sol, Wenave, Jah cose —B) sin[(x—a) -(-b)] cos (x= a) cos(x—) “* sin|(x~ a) cos (x= b)~cos(x= a) sin(x=D)) cos (=a). cos(e—B)5 U tan(x— adits ftan(x— bax] a [log| sec(x — a) |- log | sec(x—a)|]+C sin 4 sectx=a) ssee(x=b) | = any "8 exo o* ante BTS Sectaay*S ot cos(x~a) © Sinta=B)'°8 cos =) *© dx 6. Find {——__.. [CBSE 2023 (65/211 I Tazcosts =e) ‘ (sein) de Sol. Wehave, [7 oe Welare, [pe mal : de /sin?x{cosx.cosq + sinx.sina} dx cosec?x vsin‘x{sina+cosa.cotx} ” vsina + cosacotx Let sina ¥ cosacotx = t 1x Coosa) cose Ey, = ap 2Vsina+ cosa cotx 2 > Se dt Veina+cosacotx "cosa = => g Vina + cosacot x +C 7. Evaluate: [*"log( + tanaydx ICBSE 2023 (65/2/1), (AD 2011) [NCERT] Sol. Let T= (/*tog(1 +tanxydr well) a T= §togl1 + tan(F - x)]ae (By using property {* f(x)dx = ff fla— x)dx) x i tan —tanx = F/*tog)1 +4 — 1+ tan tan a Tetanx |) 2 pvt, [14 tanx +1 tans “A bist = B08 ane aia n/a I= aeeardr= GF fog? - log(t + tanx)] dx wll) Adding (i) and (ii), we get ars (toga dx =log2 ("dr = log 2px] e4 x x = flog? = T= Glog? & Find jor [CBSE 2023 (65/2/1)] ¥ aextx? Sol. Wehave, [= fe *(———>— ]ax +x)8 > x=cote 1 ex :. 1=-fe8(1—cot8 + cot?)d8 = —fe* (cosec?@- cot ®)d8 Je®(cot® - cosec?®)d8 = [e* {cot + (—cosec’8)}d8 t ‘ dx=d0 => dx =—d8 f@) —f"@) =e cot+C (fe yore fC dx=e8 f+) se tC axe HC 9. Find: [<= ae [CBSE 2023 (65/3/21 sin 3 Sol. Wehave,!= {SS ax sin 3x cos x dv a H © 1'3sin x —4sin'x = cosxde= dt s# a ata 43-427) = 1= (#3) +0(++23] For the value of A, putt = 0 1 te ax(- For the value of C, put ¢ ¥3)\.(-2v3)_ 6,3 Xi . tecx(-S)s(=28)noxd ace apse ; +See 2 = J-< Bs }+ 5 log] +tos|+-wo. Find: {x*log(x*+1)dx [CBSE 2023 (65/3/2)] Sol. Wehave, I = {x"log(x? +1)dx Letx=tand = dx=sec’9do + T= Jtan?@ log(1 + tan?6).sec?0d0 = {2 log sec. tan®6 sec@ 0 = 2flog sec. (tan?6 sec?@)d0 =i afiog se08tan?0 sec?9 40 | MPBREE® » 14n%0 08a] 9 9 2flog sec x “= u — [gx see tan x 22 *0| tan’ tan? = les m8 jog sec Li tant6 a] = tS tog seco Eta? tan?@ 2 tan*Olog sec Filet ~1)tan*0 40 3 = 2 tan°Olog sec 2 /sec"Otan?6 40 +3 jtan?0 a0 cal 2 tan89 26> = jtan’ log sec 8 tg! (sec?0-1)a0 = tan —2 tanto +2 tan 2 = jtan*@log sec8— 5 tan*® +5 tan8—58+C = log fits — 229+ 2 tant = px logy 143? — 538+ 5x 5 tan x tC [NCERT Exemplar] [sP-P =@+d)@-5) Fa+x) Se ia a- ae “le ware [CBSE (F) 2010, Dettsi 2012, 2019 (65/5/3)] Sol. Let I= fsinxsin2xsin3x dx =F) 2sinzsin2esindede = 1 J sinx(2sin2x.sin3x)dx J sinx(cosx ~ cos5x)dx [+ 2sin A sin B = cos (A ~ B)—cos (A + B)] = Fez) 2sinxcosxde— Fey! 2oinx-cosSuidx [-- 2cos sinB = sin(A + B)-sin(A-B)]Sol. 1“ Sol. 16. Sol. = Jsindxdr 4 (sinéx —sindx)dx 4 cos2x , cos6r _cosdr 6 * um 16 to eters Evaluate: /n'z+ costs 4, [CBSE Dein 2014 sin?x.cos*x sin®x+cos’x sin?x.cos'x jn?x)°+ (cos? x)? = ejiePetetat sin?x.cos?x se coe ictaccostsEcndt sin?x + cos?x)(sin*x - sin?x,cos?x+ costa) 5 pa jinteeetalents-sincoe teas sin?x.cos*x in'x ~sin?x. cos*x + cost 2. Es Ad = ftan? adr — [idx + Jcot?rde sin’x.cos’x > (sec?x -1) dx -x+ | (cosec?x-1)de => 1=Jsec?xdx+ { cosec”xdx — x —x—x+C = tanx—cotx—3x+C Evaluate: jane, [CBSE Delhi 2013; (F) 2015] sin(x—a vet 12 (EEO ay sina) Letxtast > = cos2a{ dt — f sin2a.cott dt = cos 2a.t ~ sin 2a.log | sint |+C cos 2a. (x +a) —sin 2a. log |sin ( +.a)| +C =x c08 2a +a c0s 2a - (Sin 2a) log |sin (x +a)|+C Evaluate: |! [CBSE Delhi 2009, 2019 (655/101 V5~4e* |. Let I= /——2— Put =f = ede=dt,weget ae at (Pr at-5) ° VP +202 427-9) FE ca iart( 42) Fcesin'()+c ,{sinax—4 ; Evaluate: {e*(S2o= = [CBSE Dethi 2010] fe ( Sine tet T= fe (eae) o (nee) + sin 2x =2sinx.cosx and cos 2x=1-2sin*] sin = Je*(cot2x - 2cosec”2x) dx Let fix)=cot2x-. f’ (x) =-2cosec* 2xT= fee) +f’ (dr = [ae* fix) +C=e% cot2x+C be fe) +f @)de = eX) + WF Ereake: | 8 ge [CSE (F) 2010) ae = 2) t= fag Poat2™ e- Sol tet I= Feet 18. Find: {sin [oa INCERT Exemplar] Sol. Put x=atan'@ => dr=2atanOsec'Od@ Fata? r= fsin"(y eae Seaton sec?) d0 = 2a sin” (EEF tan. sac" 848 = 2aJ sin” (sin @)tan 8. sec” 6d8 = 2a) 9. tan 8 sec*6d8 ms fem? 0d0 AS 6 ftané. sec? 88) 0} 8, tan*O— atan + a0 +C dx sing + sine 1 sing + sinde™ Vignes 2 IS sinx+2sinxcosx 19. Find: [ [CBSE Delhi 2012] Sol. Here, 1 = J. aaa eo 3 inf ier J ee f sin*x(1+2cosx) 5 (1-c0s?x) (1 +203) Letcosx=2 > -sinxdx=dz dz ade of @+20-y0 +22) “=e +22)Sol. Here, integrand is proper rational function. Therefore, by the form of partial fraction, we can write 1 Ag Big B (@+2)(-21+2) ez 712 1+2 a 1 A(1~2)(1+2z) + B(1+2)(1+22)+ C1 +2)(1-2) )( +22) (+2020 +23) = 1=A(1=2)(1 +22) + BOL +2)(1 +22) +C(1+2)(1-2) elif) Putting the value of z =—1 in (ti), we get => 1=-2A+0+0 = A=-t/2 Again, putting the value of z = 1 in (i), we get > GF => 120+820+2+0 > 1268 > B Similarly, putting the value of 2 = in i), we get = 1=0+04¢ (5)(3) = ute = ond Putting the value of A, B, Cin (i), we get 1 @+0-9d-2) ~ 20+9 “6d-| | 1 1 4 > T= Flog | 142 [+ Glog | 1-2 | -geglog | 1422 [+ 1 4 3(1 +22) 1 ~ Sia 202 6( Putting the value of z, we get > I= Flog | 1+ cosx [+ Plog | 1 cosx | Fog | 1+2c08x |+C Find: [) [NCERT Exemplar] 2 Ta 2 vont a Ef Hell let? = Seana 'eaca” | } t A B =| 03) i438 > t= A(E+3)+B(t-4) =(A+ BE + GA-AB) (On comparing the coefficient of t on both sides, we get A+Bel and 34-4B=0 = 3(1-B)-4B=0 = 3-3B-4B => 7B=3 3 then At =1 I ede (x? = 4) (x? +3)21. Evaluate: [————dx (P+ a07+9) [CBSE Delhi 2013; (F) 2015] Cal Sol. Let Ila Put x7 = t, we get 2 t (+H +) CF NETD) R t A,B _AG+9)+BE+4) Wr FED) Pea (49) = t= (A+B)t+ (A448) Equating the coefficients, we get A+B=1 and 94+4B=0 Solving above two equations, we get a. 9 (+4249) BQ? +4)” 5G? +9) f dx a 27+4)G2+9) 50x42? 5° x43? 2 AE Ft plant + ptant 4c ;_@sin®-2)cosO 22. Find: [OSB o eos J5 cos*6- asin€ Sol. Wehave = j {sin =2)cos0 5 cos*8- 4sin@ Let sinO=2 = cos@d0=dz (@z-2)dz 5-(1-z)-42 (Gz~2)dz 5-1+z2"-4z 3z-2 dz =f (@-2? @-2)? Let z-2=t => dz=dt 3+ 2)dt dt _ ptt eat ee Sa 6h 2h I ae 3z dz—2f dt P (CBSE Delhi 2016] dt git ial Blog | t [+4=Slog|#|-4.4+C Putting value of tin terms of: then zin terms of 8, we get 4 = 3logl sin0-2| e+e ae 23, Find: {3~—ydx [CBSE Delhi 20161 Va a3 Sol. Wehave Let 24, Find: [~~ — de [CBSE (North) 2016] Sol. We have, oe (2x-3)* =(ee3| 1 _ 2 Jee ee ma 1 2 a te lea mal Let2e-32t > ale=dtode= ei Z Pages § siege fa pitt > Tecpel tC Putting t = 2r-3 & 1 & Take. +C = +C 2° @x-3)? = Qx—3)? ax eon [CBSE 2019 (65/4/2)1 dx 1 cos(a—b) dx sin(x— a) cos(x—6) ~ cos(a—6)’ sin (x — a) cos (x — 6) 1 cos(a—x+x-8) ~cos(a=B) sin (x= a)eoste— 0) 1, cos((r=B)- («-a)) cos(a—b)! sin (x—a) cos (xb) cos (x — b) cos (x — a) + sin (x — b) sin (x - a) SBE NS EAD * Sn NE te sin(— 2) cos( sins-0) |, cos (x= 8) |*. coste—) 5. sine - wt Se a) oF cos (x — by | = acy Ms! sina) [tog coste—2) Je = 4 costa) 8 sin(x a) cos (x - B) ec 26. Find: [[log (log x) + [CBSE Bhubanestoar 2015, (South) 2016] 1 dx Cogn Sol. Let T= jllog(logx) + — yar (logx) Let logxst = xsd > draedt 2 Sfloges Zale =fflogi+ t+ Sear= {(tost+t Je+(-36 zl =élogt-Fd+C fb: inaeerarmaree = &P8*log (log x) - 7 eee [Put t= log x] =x log (og 2) ~apgete 27. Evaluate: f |[xt-x| dy {CBSE 2021-22 (Term-2)] ow cnengee Wan ak 0, 4y2 . So, we bane to boven i W (4,8) | wy tot) w Ga)[Topper’s Answer 2022) Sol. Adding (i) and (i), we get ari teane x = =f de=[x] =2n-0= afr ax je [x] 9° = 2n-0=22 => Tsk * 29, Evaluate: { [log (sin x) -log(2cosx)] dx Sol. Let = [log sinx —log(2cosx)]dx => = (og sinx—log2~log cosx)dx = 1 fog tans de ~tog2( ax 2 initio tanx dx —log2lx}j/” > Te? log tanxde— F10g2 voli) [CBSE 2023 (65/5/1)) [CBSE 2023 (65/1/1)] ld)Sol. a. Sol. Let 1,= [7 log tanx Ait) Using property fi foodr= i fla=xydx (Flog tan(E— x) dr = (? tog cotx de (itty Adding (i) and (i, we get 21, =f? (log tanx+ log cotx)dx = [7 tog (tanx.cotz)dx F log dx=0 = h=0 Putting f= in (9, we get [CBSE 2023 (65/1/1)1 Let I= [7 e"sinx dv Using integration by parts, we have Tesinfet de j(2S2% fetae)te T=sinxe*~ fcosxetde sinx.e*-[cosx &-J-sinx e dr] = > = 1=sinxe*-cosxe*fe* sinx de => I=(sinx-cosx)e=1 > 21=(sinx -cosx)e* == [(sinx —cosx)e 1 2 I i [[sn Fos Fe (sino ~c0s0)] =} [aoe (0-1) x spe" +1) teg/3 Evaluate: | ——* ing re «st Letl= =a oer fee (eee a (ees__tee™ es 2 (FH) Let 32 dx = dt When xelog/2 = tarot wdhal en de [CBSE 2023 (65/2/1)) (ere When x=logy/3 = t= tbe? = ebali9 =3 a1 a pest) 2 (IME=1)2*23 (DE-1 dt [loglt—11-ogl¢-+1 If.Ly, tip Li, 3-1 ke 2-1 4 8 ai) a(les3 ~'o8 35) 1 1 1\_1 3 =7(l08 3-8 3)= F083 % eins 92. Evaluate: fp, im, {CBSE 2023 (65/3/2)1 sf Six + cos!x % 109 sin! x Sol tie fae met Using property | flxidx =2 fixdae, when f(x) is even. e ° Ye gig 2 inf ge (0) 2 sin™x+ cosx Using properties f flx)dx= f fla-ardx ° cos! x i?) ——— (ii) Foal rain a Adding (i) and (ii), we have ent. 100, % sin'"x+ cos! y See dee af areata a a 33. Evaluate: {? xlog(l + 2x)dv INCERT Exemplar] Sol. Let xlog(1+2x)dr oo 2 = flog +29] Stas: =2[1log3-0]-| |x 2 jo (> tk afx?] 1} = pleas 3] 5] +3! 1 a 1 1 = Fiogs—2+1 pa 2 oglia +20 118 = Hog3— aks 1 Ss = logs - Glogs = glog334. Evaluate: | * Sol. 33, Sol. 36. Sol. dente [CBSE (AD 20181 6 1+costx Let adr G 0 1+ cos" ® _ FA@—2).sin(n—2) a 1+cos*(n-x) F4(x—2).sine pe [AS b J trent 2) Adding (i) and (ii), we get +n- © sis A@ta-asine, 5 2-4 f 280 yy 6 1 ¥ costs 6 Leos? 7 sinx an f ae "| Tscosts Leteosx=z => -sinxde=dz => sinxdr The limits are, x=052=1 yen > zs-1 122: 2nftantzy}, ] cps ahs = 2nftan“'1 - tan1(-1)] = 2r| = Ise. Evaluate: [ (cos ax —sin bx)*dx [CBSE Dethi 2015) Here, 1 [5 (cos ax-sin bx)"dx 3 T= [5 (cos* ax + sin” bx ~ 2 cos ax sin bx) dx = Le [5 cos? ax de [5 sin*bx dx — [5 2008 ax sinbx dx > 1=2 [cos ax dx +2("sinbx dx -0 ipa ous are even function while third = T= [f 2cos* ax dx + [2 sin” by dx = 1= [1 + cos 2ax) dx + [ (1 -cos2bx)dx & T= ffde+ Fcos2ax dx + [f dx — ff cos 2bx dx cs 1=2[" dx + [' cos2.ax dx - [cos 2 bx dx = roan fff > npee nen, sine Evaluate: [es ae ICBSE Delhi 2017; (AL) 2012, CBSE 2020 (65/4/1)] tet re ft ae ° 1+ c0s*xx (f—2)sin (nx) de ie Orem) x (n=x)sinx dx __ x sinxdr = anes So ecccte 8 Te costs ore n(t SME pe yx sind 1+ c0s*x 20 1+ ¢os*x Put cos x= t so that sin x dx = dt. The limits are, when x = 0, a_dt +P and x =z, =~ 1, we get [ef fende =~ [6 fle) drand ( fydr = 26) fla)ds] 2 inf = nftan“4]} = n[tan“1- tan] =x] 7" dx 37. Find: | ——&—_ [CBSE (Allahabad) 2015] 6 costx/2sin2e Hs zt Sol. Let 1= | ——%__ -f'__dx_ é costx/2sin2x 6 cos*x/22sinx.cosx alps dy a1 p__dx 2 Jsin= 2 2° costxvtanx cos*xy/ SR .cos*x aL pr sectx dx _ 1 2/4 sect x. sec?x dx 2° “Yanx 2 Vianx Lettanx=t = secxdr=dt, x20 = teOandxe> = te a+ Pat vt a) Pyarspagy alee] een - + q +4 ta pare, 2[372e1 fife 2ppey s & 2[vih+ px =1+2-$ 72 cos x 38. Evaluate: | vax [CBSE (F) 2015] oan 1¥e 2/2 cost Sol. Let I= {% ate ae in ist integrand 2/2 C08 eee et map perdtt he ede dx =-dt cost 2/2 c03 xem/2 st=n/2 = fae i + Pe ede C728 ay (7 OE ay = Pesos, a+ COSE 4 (F/2 £08% ate z tee OE em [fede] (C7 yy (om am =f ae tre [By property [' f(xyde = 2 pn ee Dcoss Tee dy = ff? cosx dx = [sinz]j = sin /2-sin = 11+ ytanx {CBSE (AT) 2011) dx . M6 1+ Jtanx @O ae a [By using property [° fayde= ff fla+¥—2)dx] 1+ /tan( dx vianx EST vane Adding (i) and (ii), we get an(l+vtanx) | (4 pak RE 2 Le Te tanay tt Hie t= bef 3 6G = usb iy 3 40, Evaluate: /{| x-1|+|x-2|+| x-3 [Ide [CBSE Delhi 2013] i sol vet r= Pty x—a +] x—2[+)x-3 [ltr= faa jde+ fea dee fjx—a) de =f|x-1|de+||x-2|de+||x-2|de+j|x- 3] dr , , ‘ [By property of definite integral] [x-120, if 1 Tha-w-G-xMx [GF fla § fla-xyaz] = (a-2eeatetde= ("20 4a" de fet oP ef a 28 n+l “nt2° nt+3 ntl nt2° +3 = (02 (1+3)—2n4 1) +3) + 1+ 1)0042) (a+ 1)(u+2)(0+3) a1 + 5nt6-2n- 81-640? +3n +2 2 (r+ 1+ 2Y(n+3) (+) +2)0+3) 44, Evaluate: | e™.sin( $+ x)éx [CBSE Dethi 2016] a Sol. Wehave I= J e*.sin(+x)dx 6 Integrating by part, we get t=fsin(S+2) 2] - BAe) F4ICBSE (North) 2016] -21t ld) 2 & distran® J foode =f fla+b—xdx Oke fF Jef a 2 5tx2 dies Adding (i) and (ii), we get = POF. J ‘T a= {OP are face [| : dx (di) =e 8 3x23 > oe its -C8)] > xtanx 46, Evaluate: fy tan dx ICBSE Dethi 2008, 2010, (AD) 2008, 2017, (F) 2010, 2013, 2074) xtanx secx+ tanx@* @ =e [-f'fendr= ff fle-vdr] Sol. tet =f (i) seex + tanx By adding equations (i) and (ii), we get tanx Multiplying and dividing by (sec x - tan x), we get x tanx (sec tanx) org ft eee) ne oc tantx = nf seextanx dx —n{sec?x dx+ (Fdx = n[secx]j —aftana]5 +1115 dx = x{" (secxtanx —tan?x)dx = x(-1-1)-0+n(2-0) = x(n -2) = UW=nn-2) > I=Z@-2[CBSE 2018] Aint tte de fo 44 Sind, sina = cot x =. (cee 4 Snax ~ ae iL Pak Abo (vainn ~ cin)” = saint. 4 onthe — aetna cern. = " t= sfinax=t™ Binet = tot sbivih | tuo | 4209 t= -1 Hm] 0 lle le + 8084) st Pe a 1. Evaluate: { Sol. > srt AQx+4)+B \ t5q,{ 2000- leg, | BOU=. | sy seal — ea Ses | 6 | alee} +| | a uw j= ae geet jal) oe_ feng? ae [Topper’s Answer 2018) Long Answer Questions [CBSE Delhi 2011; (AD) 2010] We can express the N’ as Sr+3e AL (7 +4e+10) +B => Sr#3=2Ax+(44+8)Equating the coefficients, we get 2A=5 and 4A+B=3 5 5 A=Z 2 4XZ+B=3 + B=3-10=-7 set3=Serta+(-7) 5 Sarta-7 1-2 vx? +4x+10 Vx? +4x+10 Let 2 44x410=f=> (2x 44)dx=dt 1, =2yx?+4x+10+C, dx ¥x+242+2—-4+10 © y(x+2)7+(/6)" Again, 1, = log| (x +2) + Vx? + 4x +10 |+C, Putting the value of J, and J, in (i), we get Soy r 5 T= 3 xayx? +4e+10 -7og | (¢+2)+ V2? +4x+10 |+(7¢,-76,) = 5yix*+4x+10 —Zlog | (x+2)+ Vx? +4x +10 |+C, where C=! C-7C,) 2. Evaluate: {——7* dx [CBSE Delhi 2011] (P+t+3) Sol. Let =z > 2xdx=dz 9 _gewip _ te _. G@eieea eer Using partial fraction Let i a _) (@+1E+3) "241 243 1 Ae+3)+ BG +1) G+DE+3) G+DE+S) > 1=A(e+3)+Be+1) > 1=(A+B)z+(3A+B) Equating the coefficient of z and constant, we get A+B=0 tii) and 3A+Be1 lili)Subtracting (ii) from (ft), we get zaz1 = a= and s=d 2 Putting the values of A and B in (i), we get 1 1 1 G+E+3)” 2E+1) 2e+3) f 2x dx G24 1)G? +3) 1 1 dz “I (ery Term) gel 2/243 log |2#3 |+C= Slog | 22 +1 | Flog | x7+3 |+C Note: logm + logn = logm.n and logm—logn = logm /n [CBSE (AD) 2014] Sol. Let I= ‘ \Girsere Now, we can express as 4 estos x42 > kH2=AQr+5)+B > x42 =2Ar+(GA+B) Equating coefficients both sides, we get 2421,5A+B=2 0 > OA 1 1 xt2=5Qr+5)-5 oo ®ae (+3) = log |(x+Z) +4546 tg Putting the value of J and lyin (), we get 1 3 fleai(x+ Ves 1 5\.F 1 = Vx? + 5x46 ~ Flog | (x+5)+ V3? +5446 [+ 56, 1, I= 50x +5x+6+C) Je vs? +5e+6 [+ C} mee 5), ace = VP FEREG — Hog |(x+3)+ VP Br+6 [HC (Here, C= 1 4, Evaluate: { ———, dr [CBSE (Al) 2014) sin'x+ sin? xcos'x+ cos 1 sin*x + sin’x.cos?x + cos*x Dividing N’ and D’ by cos* x, we get Sol. Let I= dx 4, sectx 1" ene tan'x+tan'xe+1 Put zstanx > dz=sec?xdx G42 d | ge2el [NCERT Exemplar] Sol. Put x=acos20 = dx=—a.sin20.2.40 ‘eect asin 2040 [Sesh sin 200 = 2a) / 25°57 sin 208 = si 4a cos” 0d0 =—2af (1+ cos 26)d0 i ttl [| a “0s atpy t-tePutting /x = cos, ie, x=cos?@ => 6 = cos x and dx =~ 2cos@sinOd®, we get /1-cos8 T= IW Te eoe8 28in Geos6) a8 { asin? ® ain =-2f | 4 (sin 8 cos 8) d9 = — 2f—Basingcosheos 00 e055 cos 5 =2) sin? Leos -2J (1- cos 8) cos .d® =-2f (cos @- cos”). d0 2) cos 040 + {2cos?6..d8 =-2sin 8 + (1 + c0s28).d8 sin 20 2 =—2sinO +/1.40 + J cos 20.40 =-2sin8 +0 + + 2y/1— cos? A.cos@ -2y/1— cos? + 9+ = ~2VT=x teos* Vxt vx VI-x+C 7 2 7. Find: |———*~—_dx (sinx + cosx) 2 ds sou rae] 2 pape _ Fg (xsinx + cosx) (xsinx + cosx)? Cos, Integrating by parts, taking <2 as the first function and Ge a ‘as the second function, fsinx + cos2) swe get x xeosx dy x xcosx 1 dal ag a Now, let us first evaluate [SS (sins + cosx) Putting (x sin x + cos x) =, then (sin x +x cos x—sin x)dx =dt ie, x cos xdx = dt, we get xcosx ao. 1 j_ bron aaa fa ——* _. (xsinx + cosx) ? (xsinx+cosx) 4 + a Fence; f=! cosa asins 1g fences 1 Cosi (xsinx + 6083) costx (sine * c0s2) x a bd. = et fcc te § tant tr Gamay cos x(@ sin x+ cos x) © a cet rsin?x + sinxeosx | _ =x(1-sin* x) +sinxeose cos x(x sin x + cos x) cos x(x sin x + cos x) _ £08 (sin xx 008 x) +c * cos x(x sin x* cos x) Side _(sinx-xcosx) , c (xsinz + cosx) (xsinx + cosx)?sin?x 8. Evaluate: (°"—""* — ax [CBSE Panchkula 2015; (South) 2016] sinx+ cos 2 _ pe__sintx . Sol. Let I= =a i) #12 ¢g32. osx 1S Sa ® 9 cosx+sine Adding (i) and (i), we get “2 sin?x + cos" 4 dx sinx-toosx “ g sinx+cosx 1 de = aa! v2a Ty anvein™ cosx.cos ++ sinzsin= 4 4 ix [1s 208 (A-B) = cos A cos B + sin A sin B] tog(/2++1)—tog( see —tan)] i 3 -1yj= = yy llog(/2 + 1)-log(v2 - D1 “2 toe[ 44] “bol E22 i 2 ai elo log(/2+ 1)? = F5logiv2 +1) 1 Hence, = —log(/2*1 nce, T= ry log(y2 +1) 72_xsinxcosx 9. Evaluate: [ee [CBSE Delhi 2011, 2014; Sample Paper 20171 0 sintx+cos*x Sol. Let ade sin'x + cos" By P: . ty Property LO fladde = fF fla x)ax| sp (5—2) cos. sin > rs | -4§— ote a cost + sin’x? cosx.sinx , 7? xsinz.cosx & dx—f ent sint+costx” @ sintx+costr is 2 gi cosx.siny n¥/2_sinx.cosx > OSES te > 21a Ef SRO ty sintx + costs 29. sintx+coste sin x.€05 x 4 aP cost ‘s =3) aa [Dividing numerator and denominator by cos* x] @ tan’x+1 __t f? 2tanx.sec*x “2x25 14 (tan2x? Tet tan?x=2; 2tan x. sec? xdx =dz = x 10. Evaluate: [*———* dx [CBSE (A1) 2009; (F) 2014] a cos*x +B sin®y : x Sol. Let He) Peotz+ Pants ™ m0} t= f= tz f rayde =f fear 6 a cos*(a—x) +b? sin (xa) be [fedex | fea] 1 Rox ay i) 0 @°cos*x + bsin?x Adding (i) and (ii), we get a= f= 3 eos +P sins [Divide numerator and denominator by cos” x] Pa 6 J f(x) = 2) fx) atx] Put btanx - ab 11, Evaluate the following: ("| xeosxx | dv [CBSE (F) 2010; Patna 2015; (Central) 2016] Sol. "| xcosnx | dx x Asweknow, cosx=0 = x=Qr-Ipn|ezcos nx =0 = 1 For 0<1<5,1>0 then - cosmr>0 => —-xcosnx>0 1 3 For 50then cosmx<0 = = xecosnr<0 AR? | xcosnx | dx = ff”? xeosmxde + [29 (— xcosmx)dx sinnx}? a, si ? pr sinax =lx fp" 1. +i x hya x ei dx soosmx (ae*-zs)-(-ae ae) (CBSE (F) 2016) a ST gnasina aay 7 x nf —* _s | Teeinasin =f" faeces d xi = 0 1+sina.sinx ax 2tans 1+sina- z 1+ tan? > 5 1 + tan?) 5 set f hem xf ———42__«x © tetan?S+2sinatane © tan? +2sina.tand+ 1+ tan? +2sina.tan tan? 5 +2sina.tan5+1 = sec? Fde= Ut; x= 0 = f= Oandrex = fc edt a distal ae 0 P42sinat+1 f—___# _. ¢ P+2sinat + sin?a—sin?a+1 _ a ‘a (f+sina)? +(1-sin?a) Let tan at #. 2 0 (sina) + costa am oa tan5+sina cosa Phe testa x = cosal'®" cose = =—™ IE tant = aglF-tartaney] x(x-2a) 2cosa13. Evaluate: f sin 2x tan"(sin x)dr [CBSE 2023 (65/5/11 é 4 4 Sol. I= f sin(2x)tan7 (sin x)dx = f tan“ (sin x)sin (2x)dx a é [tan (sinx) sin(2x)de = tan” (sinx) { sin(2x)dx — 5 {fe ltan (sina) {sin 2rd ax -cos2x 2 tan (sina) x (=) 4 3) peambe eos ( Jax sin? ne Fen (sinx) cos(2a) + $f =A cost [+08 2x=1-2sin?x] Putsinx=u => — cosxdx=du si 1 )1-207 pian (sina) eos(20)+ 5 /- rd = 2u+3tanty -2sinz +3 tan (in x)+C Now, tan" (sinx) sin(2x)dx 1 1 = = plan sinx) cos(2x)+-5[-2sinx+3tan(sinx) +C] % 2 Te | flan ins) cos(2e) sin + tan sina] * ° 1 e(sin™ 1) gin + Stan (sin® =[[ Gee (sin) cos(2% %) -sin + $ tan" (sin )} tan(sin0) cos(0)—sin(0)+ Stan sino] = [lee 3 tant ee = [Ram 1) eos(n) -14F tan «}-o] [eo tar iste frtan*a)=Z and cose