TO BE SUBMITTED SHEET-
Strenthening your Concept TBS-Integration-Last
maths by rupesh k jha.....9864030380
www.infinity4maths.com....7086055586
ASKING DOUBTSARE YOUR FUNDAMENTAL RIGHT,
By Taking factor common
(ax 2 b) dx 5 x 4 4 x5
Q1 x c2 x 2 (ax 2 b) 2
Q2
x 5
x 1
2 dx
By Adjusting ex
1 sin x cos x
2 e x sin x
Q.3 Let F(x) = dx and F(0) = 1 then F(x) is equal to
(A) ln 1 e x sin x 1 (B) ln 1 e x sin x
1 1
(C) ln (1 e x sin x ) 1 (D) ln (1 e x sin x ) 2
2 2
Integration by Parts and Cancellation
coec 2 x 2005
Q4 cos 2005 x dx
Diff Both Side
Q.5 If x
2
·e 2 x dx = e–2x(ax2 + bx + c) + d, then
1 1 1 1 1 1
(A) a = – ,b=– ,c=– (B) a = – , b = – , c =
2 2 4 2 2 4
1 1 1
(C) a = – , b = – 1, c = – (D) a = 1, b = 1, c = –
2 2 2
3e x 5 sin x 10 cos x
Q.6 Let e x 4 sin x 3 cos x dx = m x + n ln (ex + 4 sin x + 3 cos x) + C,
where C is constant of integration. Find the value of (m2 + n2).
By integrating Both Side
3
Q.7 Let f be a twice differentiable function on R and satisfying f "(x) = x 1 2
2
for all x R.
If f ' (0) = 0 and f (1) = 2 + 1 , then f (0) is equal to
(A) 0 (B) 2 (C) 3 (D) 5
Misc
1 x dx 1
Q8 Prove that : 3 =– 2 tan
–1 x 1 + C
1 x x x2 x x
tan x 1 2 tan 2 x tan x
tan x 2 dx =tan–1 + ln C
2
Q9 Prove That
2 tan 2 x 2 2 tan 2 x tan x
7
cosec 2 x 1 cosec x cot x 2
If = cosec x cot x C where C is constant of
2
Q.10 dx
9
cosec x cot x 2 11
integration and N, then is equal to
7
(1) 5 (2) (3) 10 (4) 7
2
3 cos x 2
Q.11 (3 2 cos x ) 2 dx is equal to
sin x 2 cos x sin x sin x
(1) C (2) C (3) C (4) C
3 2 cos x 3 2 cos x 3 2 cos x 2 3 cos x
Integration in Pair
ex e x
Q.12 Let I = 4x dx and J = 4 x dx
e 1 e 1
Then for any arbitrary constant C, match the following
Column-I Column-II
1 e2x 1
(A) I (P)
2
tan–1 2e x C
1 e 2 x 2e x 1
(B) J+I (Q) ln C
2 2 e 2 x 2e x 1
1 1 e 2 x 1 1 e 2 x 2e x 1
tan
(C) J–I (R)
2 2 2e x 2 ln e 2 x 2e x 1 C
1 1 e 2 x 1 1 e 2 x 2e x 1
tan
(S)
2 2 2 e x 2 ln e 2 x 2 e x 1 C
By Forcing Integrals
x2
Q13 ( xsinx cosx)2 dx
Derived Substitution
1 x 1
1
Q14 tan 2 tan dx
1 x 1
PART-11
1
Q.1 If f (x ) dx = g (x), then f ( x ) dx is
(A) x f–1 (x) + C
(B) f g 1 ( x ) C
(C) x f 1 ( x ) g f 1 ( x ) C 1
(D) g ( x ) C
1 x
ln ln
1 x dx Ans.
Q.2 1 x2
Q.3 (sin x)11/3 (cos x)1/3dx
Q.4 Let f be a polynomial function such that for all real x f (x2 + 1) = x4 + 5x2 + 2, then f (x ) dx is
x 3 3x 2 x 3 3x 2
(A) – 2x + C (B) + 2x + C
3 2 3 2
x 3 3x 2 x 3 3x 2
(C) – 2x + C (D) + 2x + C
3 2 3 2
3x 4 1
Q.5 Primitive of w.r.t. x is :
( x 4 x 1) 2
x x x 1 x 1
(A) +c (B) +c (C) +c (D) +c
x 4 x 1 4
x x 1 4
x x 1 4
x x 1
cos x sin x 1 x
Q.6 If e x sin x x
dx = ln f ( x ) + g(x) + C where C is the constant of integration and f (x) is positive,
then f (x) + g (x) has the value equal to
(A) ex + sin x + 2x (B) ex + sin x (C) ex – sin x (D) ex + sin x + x
dx
Q.7 equals
2
x x 1 5
4/5
5
1 x 5 5
1 x 5
5
1 x 5 5
1 x 5
(A) c + (B) c (C) c (D) c +
4x x 5x x
(x
3
Q.8 If 2 x 2 5)e 3x dx = e3x (Ax3 + Bx2 + Cx + D) then the statement which is incorrect is
(A) C + 3D = 5 (B) A + B + 2/3 = 0 (C) C + 2B = 0 (D) A + B + C = 0
dx 1 x q
Q.9 Let 2008 = p ln r +C
x x 1 x
where p, q, r N and need not be distinct, then the value of (p + q + r) equals
(A) 6024 (B) 6022 (C) 6021 (D) 6020
Q10
Evaluate sin7 x cos x dx
5
1 1
Q.11 If ( x 9 x 6 x 3 ) (2 x 6 3x 3 6) 3 dx = (2x9 + 3x6 + 6x3)B + C, where C is integration constant then AB
A
is equal to
(1) 32 (2) 16 (3) 8 (4) 4
Answer
1
ax 2 b x1 x5
1) sin k (2) C– 5 or C + 5 (3)a 4)-
cx x x1 x x 1
(cotx)(secx)2005+c
5)a 6)5 7)b 10)4
11)3 12)A) R ; (B) P ; (C) Q
xsecx 4 5/4
13) c 14) x +c
( xsinx cosx 5
1 1 x 1 x 1 x
1)c 2) ln 1 x ·ln ln 1 x ln 1 x + C
2
3)
3 1 4 tan 2 x +C 4)a 5)b
8(tan x ) 8 / 3
6)b 7)b 8)c 9)c
10)-(cos6x)(cosx)6/6+c 11)1
