SAINT MARY’S ANGELS COLLEGE OF PAMPANGA
COLLEGE OF ENGINEERING
CIVIL ENGINEERING DEPARTMENT
MATHEMATICS, SURVEYING AND TRANSPORTATION ENGINEERING
INTEGRATION FORMULAS
Algebraic, 1. ∫ a du=a∫ du=au+C
Exponential, and
2. u au
Logarithmic ∫ a du= ln a
+C , a> 1, a ≠ 1
Functions 1
3. ∫ u n du= n+1 un +1+ C for n≠−1
4. ∫ eu du=eu +C
du
5. ∫ u−1 du=∫ u
=ln|u|+C
6. ∫ lnu du=u ln|u|−u+C
Trigonometric 7. ∫ sin u du=−cos u+ C
Functions 8. ∫ cos u du=sin u+C
9. ∫ tan u du=ln|sec u|+C
10. ∫ cot u du=ln|sin u|+C
11. ∫ sec u du=ln|sec u+ tanu|+ C
12. ∫ cscu du=ln |csc u−cot u|+C
¿−ln |csc u+cot u|+ C
13.∫ sec u tan u du=sec u+C
14.∫ cscu cot u du=−csc u+C
15.∫ sec ² u du=tan u+C
16.∫ csc² u du=−cot u+C
Inverse du u
17.∫ =arcsin +C
Trigonometric
2
√ a −u ² a
Functions du 1 u
18.∫ = arctan + C
√ a +u ² a
2 a
du1 u
19.∫ = arcsec +C
u √ u −a ²2 a a
SAINT MARY’S ANGELS COLLEGE OF PAMPANGA
COLLEGE OF ENGINEERING
CIVIL ENGINEERING DEPARTMENT
20.∫ arcsin u du=u arcsin u+ √ 1−u2 +C
21. ∫ arctan u du=u arctan u−ln √1+u ²
Hyperbolic 22.∫ sinh u du=cosh u+C
Functions 23.∫ cosh u du=sinh u+C
24.∫ tanh u du=ln cosh u+C
25.∫ coth u du=ln |sinh u|+ C
26.∫ sec 2 u du=tanhu+ C
27.∫ csch ² u du=−coth u+C
28.∫ sech u tanh u du=−sech u+C
29.∫ csch u cothu du=−csch u+C
du u
30.∫ =sinh −1 +C
√ u +a 2 z2
du u
31.∫ =cosh−1 +C , u> a>0
2
√ u −a ² a
du 1 u
32.∫ = tanh−1 +C ,u 2< a ²
a −u ² a a
2
du −1 u
33.∫ = coth −1 +C , x 2 >a ²
a −u ² a a
2
Other Functions du
34. ∫ =ln |u+ √ u ² ± a ²|+C
√ u2 ± a ²
35. ∫ a2du 2
1
= ln
−u 2 a | | a+u
a−u
+C ,u 2< a ²
36. ∫ u du2
1
= ln |
−a ² 2a
u−a
u+a |
+ C , u >a ² 2
2 2 u 2 2 a² u
37.∫ √ a −u du= √ a −u + arcsin +C
2 2 a
u 2 2 a²
38.∫ √ u ² ± a ² du=
2
√u ± a + 2 ln |u+ √ u ² ± a ²|+C
Integration by 39.∫ udv=uv−∫ v du
SAINT MARY’S ANGELS COLLEGE OF PAMPANGA
COLLEGE OF ENGINEERING
CIVIL ENGINEERING DEPARTMENT
Parts
Trigonometric Some integrations may be simplified with
Substitution the following substitutions:
1. If an integrand contain √ a−x2,
substitute x 2=a sin ² θ
2. If an integrand contain √ a+ x ²,
Substitute x 2=a tan ²θ
3. If an integrand contain √ x 2−a,
Substitute x 2=a sec ²θ
More generally, an integrand that contains
one of the forms √ a−b x 2 , √ a+bx ² ,∨ √b x 2−a
but no other irrational factor may be
transformed into another involving
trigonometric functions of a new variables
follows:
For Use To obtain
a
√ a−bx ² x 2= sin ² θ √ a √ 1−sin ² θ=√ a cos θ
b
2 a
√ a+bx ² x = tan ² θ √ a √ 1+ tan ² θ=√ a sec θ
b
a
√ b x 2−a x 2= sec ² θ √ a √ se c 2 θ−1=√ a tanθ
b
Wallis Formula π /2
∫ sin m θ cos n θdθ=¿ ¿ ¿
0
where:
α =π /2 when both m and n are even
α =1 if otherwise
m∧n=¿ positive integer, not equal to 1
PLANE AREAS
Using Vertical Strip
SAINT MARY’S ANGELS COLLEGE OF PAMPANGA
COLLEGE OF ENGINEERING
CIVIL ENGINEERING DEPARTMENT
x2
A=∫ ( Y U −Y L ) dx
x1
Y U =f ( x ) ; Y L=g (x)
Y U =g ( x ) ;Y L =f (x)
Using Horizontal Strip
Y2
A=∫ ( X R− X L ) dy ¿ ¿
Y1
X R=g ( y ) ; X L =f ( y )
By Polar Coordinates
θ2
A=1/2∫ r 2 dθ
θ1
SAINT MARY’S ANGELS COLLEGE OF PAMPANGA
COLLEGE OF ENGINEERING
CIVIL ENGINEERING DEPARTMENT
PROBLEM SETS
PROBLEM 1 Integrate: ( 7 x 3 +4 x 2 ) dx
7 x3 4 x 2 7 x4 4 x3
+ +C + +C
A. 3 2 C. 4 3
7 x4 4 x2 4 4 x2
+ +C 7x − +C
B. 4 5 D. 2
PROBLEM 2 4 dx
Evaluate:∫
3 x+ 2
1
ln ( 3 x+ 2 )+ C
A. 4 ln ( 3 x+2 ) +C C. 3
4
ln ( 3 x+ 2 )+C
B. 3 D. 2 ln ( 3 x +2 ) +C
2
PROBLEM 3 Evaluate the integral of e x +1 2 x dx .
2
e x +1
+C 2
A. ln 2 C. e x +1 +C
B. e 2 x +C D. 2 x e x +C
PROBLEM 4 dx
Evaluate ∫ ln x x √¿ ¿ ¿ ¿ ¿
A. arcsec ¿ ¿ C. ln √ ¿ ¿ ¿
2
¿¿
B. 3 D. arcsin ¿ ¿
PROBLEM 5 Evaluate ∫ arctan x dx .
A. arctan x−ln √1+ x 2 +C C. x arctan x−ln √ 1+ x2 +C
B. arctan x +2 ln(1+ x 2)+C D. x arctan x−2 ln ( 1+ x2 ) +C
SAINT MARY’S ANGELS COLLEGE OF PAMPANGA
COLLEGE OF ENGINEERING
CIVIL ENGINEERING DEPARTMENT
PROBLEM 6 Evaluate the integral of x cos 2 x dx with limits from
0 ¿ π / 4.
A. 0.143 C. 0.114
B. 0.258 D. 0.186
PROBLEM 7 2 2y
Evaluate ∫∫ ( x 2 + y 2 ) dx dy.
1 0
A. 35/2 C. 17/2
B. 19/2 D. 37/2
PROBLEM 8 1 z y
Evaluate ∫∫∫ dx dy dz.
0 0 0
A. 1/3 C. 1/2
B. 1/4 D. 1/6
PROBLEM 9 Find the area of the region bounded by the curves
12 x
y= , the x−axis , x=1, and x=4.
x 2+ 4
A. 4 ln 6 C. 6 ln15
B. ln 24 D. 6 ln 4
PROBLEM 10 Find the area of the region bounded by one loop of
the curve x 2= y 4 ( 1− y 2) .
A. π sq . units C. ( π /4 )sq . units
B. ( π /2)sq . units D. ( π /8) sq . units
