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100 Integrals

Mahayodha Academy
(100 Integral )
(Great for calc 1 and calc 2 students)
Video: https://youtu
...
Gaikwad Nitin
IIT-JEE Main Notes

1

Mahayodha Academy

Indefinite Integration
(Step by Step Solution )
∫
(Q1
...
)

Aside
cos (2x) = cos2 x − sin2 x
= (cos x − sin x)(cos x + sin x)

= sin x + cos x + C

∫

x2 + 1
dx
− x2 + 1
∫
1 + x12
=
dx
x2 − 1 + x12
∫
du
=
u2 + 1

(Q3
...
)
=
=

2

∫

(x + ex ) dx
(x2 + 2xex + e2x ) dx

1 3
1
x + 2xex − 2ex + e2x + C
3
2

IIT-JEE Main Notes

Aside
1
1
= x2 −2 + 2 +1
x2
x
1
= (x − )2 + 1
x
1
Let u = x −
x
1
du = (1 + 2 ) dx
x

x2 − 1 +

Aside
D
I
+ 2x ex
− 2 ex
+ 0 ex

2

Mahayodha Academy

∫
(Q5
...
)

1 ln | sin x − 6|
+C
7 ln | sin x + 1|

=

Let u = sin x
du = cos x dx
u2 − 5 u − 6 = (u − 6)(u + 1)
u − 6 = 0 when u = 6
sub u = 6 into u + 1 = 7
u + 1 = 0 when u = −1
sub u = −1 into u − 6 = −7

∫

1
√
dx
ex
∫
1
=
x dx
e2
∫
x
= e− 2 dx

(Q7
...
)

Aside
√
Let u = ex − 1
ex
du = √ x
dx
2 e −1
√
2 ex − 1
dx =
du
ex
u 2 + 1 = ex

∫

1
√ dx
x+ x
∫
1
= √ √
dx
x( x + 1)
∫
√
1
2
x du
= √

x × u
√
= 2 ln ( x + 1) + C

Aside
∫

(Q9
...
)

5

−1

|x − 3| dx

=8+ 2
= 10

A=8
−1

IIT-JEE Main Notes

y = |x + 3|

y

4

4

2
3 2 5

4

2
x

Mahayodha Academy

∫
(Q11
...
)

√
= − sin−1 x 1 − x2 + x + C

D
sin−1 x

+
−

...
)

Aside
sin 2x = 2 sin x cos x
Let u = sec x + tan x
du = sec x tan x + sec2 x dx
= sec x(tan x + sec x) dx

= ln | sec x + tan x| + C

IIT-JEE Main Notes

5

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∫
cos2 (2x) dx
∫
1
=
(1 + cos 4x) dx
2
1
1
= (x + sin 4x)
2
4
1
1
= x + sin 4x + C
2
8

(Q14
...
)
3
x +1
∫
∫
1
1
1
x−2
dx −
dx
=
3
x+1
3
x2 − x + 1
∫
∫
1
1
1 1
2(x −1 − 3)
=
dx −
dx
3
x+1
3 2
x2 − x + 1
∫
1
2x − 1
3
√
) dx
= I1 −
( 2
−
1
2
2
6
x − x + 1 (x + ) + ( 3 )2
2

1=

0+1
1
1=
+C
3
2
C=
3
use x = 1

2

x− 1
1
1
1 2
= ln (x + 1) − (ln (x2 − x + 1)) + √ tan−1 ( √ 2 )
3
3
6
2 3
2

+

B×0+ C
02 − 0 + 1

1
B × 1 + 23
1
3
=
+
(1 + 1)(12 − 1 + 1)
1 + 1 12 − 1 + 1
1
1
2
=
+B+
2
6
3
1
B=−
3
d 2
(x − x + 1) = 2x − 1
dx
1 3
x2 − x + 1 = x2 − x + +
4 4

1
1
1
1
2x
=
ln (x + 1) − ln (x2 − x + 1) + √ tan−1 ( √ − √ ) + C
3
6
3
3
3

IIT-JEE Main Notes

1
3

6

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∫
x sin2 x dx
∫
1
=
x (1 − cos(2x)) dx
2
∫
∫
1
= ( x dx + −x cos (2x) dx)
2
1
1
1 1
= ( x2 − x sin (2x) − cos (2x))
2 2
2
4
1
1
1
= x2 − x sin (2x) − cos (2x) + C
4
4
8

(Q16
...
)
x
∫
= x2 + 2 + x−2 dx
=

1
1 3
x + 2x − + C
x
3

∫

3
dx
x2 + 4x + 29
∫
1
dx
=3
(x + 2)2 + 52

(Q18
...
)
∫

cos5 x
dx
sin5 x
∫
( 1 − sin2 x)2 cos x
dx
=
sin5 x
∫
(1 − u2 )2
du
=
u5
∫
= u−5 − 2u−3 + u−1 du

=

= −

Aside
cos4 x = (1 − sin2 x)2
Let u = sin x
du = cos x
1
= csc x
sin x

1
csc 4 x + csc 2 x + ln | sin x| + C
4

IIT-JEE Main Notes

7

Mahayodha Academy

∫
(Q20
...
)

∫

sin3 x cos2 x dx

( 1 − cos2 x) cos2 x sin x dx
∫
= − (1 − u2 ) u2 du
∫
= − (u2 − u4 ) du

Aside

=

sin x = sin2 x × sin x
3

= (1 − cos2 x) × sin x
Let u = cos x
du = − sin x dx

1
1
= − cos 3 x + cos 5 x + C
3
5

∫

1
√
dx
x2 x2 + 1
∫
x−3
√
dx
=
1 + x−2
∫
1
1
√ du
=−
2
u
∫
1
1
=−
u− 2 du
2
√
1
= − 1+ 2 +C
x

(Q22
...
)

∫

=
∫
=

sin x sec x tan x dx
tan2 x dx
(sec2 x − 1) dx

Aside
sin x
sin x sec x =
cos x
= tan x

= tan x − x + C

IIT-JEE Main Notes

8

Mahayodha Academy

∫
(Q24
...


sec3 x dx = sec x tan x + ln |secx + tan x|
=

D
sec x
sec x tan x

I
sec2 x
tan x

tan2 x = sec2 x − 1

1
1
sec x tan x + ln |secx + tan x| + C
2
2

Aside
9x = ( 3x)2
2

∫

1
√
dx
x 9x2 − 1
∫
1
√
dx
=
1
sec2 θ − 1
3 sec θ
∫

1
1
θdθ

tan
θ
sec
=




1
3
θtan θ 
sec
3
∫ 
= 1 dθ

Let 3x = sec θ
1
x = sec θ
3
1
dx = sec θ tan θ dθ
3

(Q25
...
)

IIT-JEE Main Notes

dx = 2u du
D
2u
2
0

9

I
cos u
sin u
− cos u

Mahayodha Academy

∫
(Q27
...
)
∫ √
=
(x + 2)2 + 32 dx
∫ √
=
( 3 tan θ)2 + 32 (3 sec2 θ) dθ
∫ √
= 3 ( tan θ)2 + 1 (3 sec2 θ) dθ
∫
= 9 sec3 θ dθ

x2

Aside
√
+ 4x + 13 =
x2 + 4x+ 4 +9
√
=
(x + 2)2 + 32
Let x + 2 = 3 tan θ
dx = 3 sec2 θ dθ
√
√
tan2 θ + 1 = sec2 θ
= sec θ
∫
recall sec3 x dx
1
1
sec x tan x + ln |secx + tan x|
2
2
x+2
tan θ =
3
=

1
1
= 9( sec θ tan θ + ln | sec θ + tan θ|)
2
2
√
√
2 + 4x + 13
x
x2 + 4x + 13
9
(x + 2) 9
x+2

=
×
+ ln |
+
| + C1
2
2
3
3
3
3


√
9 √
1
= (x + 2) x2 + 4x + 13 + ln ( x2 + 4x + 13 + (x + 2)) + C2
2
2

2

+
√ (x θ

2
2)

+

3
C2 = 3 + C1

∫
∫
5

(Q29
...
)

(x − 3)9 dx

3

∫

u=2

u9 du

=
[

u=0

]2
1 10
=
u
10
0
1 10
1 10
=
2 −
0
10
10
1 10
1 10
=
2 −
0
10
10
512
=
5

∫

Let u = x − 3
du = dx

1

√

dx
3
x − x2
∫
1
=
√ √
√ dx
x 1− x
√
∫
x
−2
= √ √
du

x
u
∫
1
1
=−2
2u− 2 +1= 2 du

(Q31
...
)

√

=2 sin−1 ( u)
√
= 2 sin−1 ( x) + C

IIT-JEE Main Notes

Aside
Let u =

√

x

1
√ dx
2 x
√
dx = 2 x du
du =

11

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∫
(Q33
...
) √ dx
x
∫
√
1
=2 x ln x −2 √ dx
x
√
√
= 2 x ln (x) − 4 x dx + C

−

I

− 21

x√
2 x

√
√
x
1
× 2 x = −2
x
x
√

x

= −2 √ √

x
 x
2
= −√
x

∫

1
dx
+ e−x
∫
1( ex )
dx
=
ex (ex + e−x )
∫
ex
dx
=
x
( e )2 + 1
∫
u
du
=
2
u +1

(Q35
...
)

∫

log2 x dx

ln x
dx
ln 2
∫
1
=
ln x dx
ln 2
1
=
( x ln x − x) dx
ln 2
x ln x
x
=
−
dx
ln 2
ln 2
x
= x log2 (x) −
+C
ln 2
=

Aside
+∫
−

D
ln x
1
x

I
1
x dx

Aside
∫
x3 sin(2x) dx

(Q37
...
)

IIT-JEE Main Notes

Aside
Let u = 1 + x3
du = 3x2 dx
1
du = x2 dx
3

13

Mahayodha Academy

∫

1

(Q39
...
)

√

y=

x2 − 1 dx

1

=

√

t
2

(2)

2

x

1

or
=

3−

cosh

−1

y=

√

√
sinh−1 ( 3)
3−
2

IIT-JEE Main Notes

√

√

x2 − 1

22 − 1 =

2

√
√
22 − 1 = 3
√
t
1
×2× 3−
Area of triangle =
2
2 √
cosh t = 2, sinh t = 3
√
t = cosh−1 (2), t = sinh−1 ( 3)
√
√
√
sinh−1 ( 3)
cosh−1 (2)
or 3 −
Area = 3 −
2
2

When x = 2, y =

14

Mahayodha Academy

√

3

∫
(Q41
...
)

∫
(Q43
...
)

1
√
dx
2
x +1

= sinh−1 x + C

Aside
d
1
(sin−1 x) = √
dx
1 − x2
d
1
(sinh−1 x) = √
dx
1 + x2

∫

(
)
√
ln x + 1 + x2 dx
∫
x
−1
√
=x sinh x −
dx
1 + x2
√
= x sinh−1 x − 1 + x2 + C

Aside

(Q45
...
)

∫

tanh x dx

sinh x
dx
cosh x
∫
du
=
u

=

Aside
Let u = cosh (x)
du = sinh (x) dx

= ln | cosh x| + C

∫
(Q47
...
)

= x tanh−1 x +

(Q49
...
)
0

4

= 10

3

2
2

1

1
0

IIT-JEE Main Notes

1

2

3

4

17

5

x

Mahayodha Academy

∫
(Q51
...
)

= −

1
1
+C
15 ( 5x + 2)3

Aside
Let u = 5x − 2
du = 5 dx
1
du = dx
5

∫

(
)
ln 1 + x2 dx
∫
=x ln (1 + x2 ) −2

Aside

(Q53
...
)

= −

x4

1
ln | 1 + x−3 | + C
3

IIT-JEE Main Notes

Aside
x4 + x = x4 (1 + x−3 )
Let u = 1 + x−3
du = −3x−4 dx
du
dx =
−3x−4

18

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∫

1 − tan x
dx
1 + tan x
∫
sin x
1 − cos
x
=
sin x dx
1 + cos
x
∫ cos x−sin x
cos x
=
cos x+sin x dx
cos x
∫


cos x − sin x
cos
x

=
dx
×


cos x + sin x
cos
x

∫
cos x − sin x
=
dx
cos x + sin x
∫
du
=
u

(Q55
...
)

x sec x tan x dx

= x sec x − ln | sec x + tan x| + C

+
−
+

D
I
x
sec x tan x
1
sec x
0 ln | sec x + tan x|

Aside

∫

sec−1 x dx
∫
=x sec−1 x −
∫
−1
=x sec x −
∫
=x sec−1 x −
∫
=x sec−1 x −

(Q57
...
)

∫
(Q59
...
)

1

√

−1

∫

=2

1

2

4−

x2

π
3

dx
2

√
4 − x2 dx

√
π
1
1
=2( × 1 × 3 + 22 × )
2
2
6
√
2π
= 3+
3

IIT-JEE Main Notes

−2 −1

3

1
√

0

√

1

3
2

x

1
π
θ = sin−1 ( ) =
2
6
1 2
Area of Sector = r θ
2

20

Mahayodha Academy

∫ √
(Q61
...
)

=

1 x3
e +C
3

∫
(Q63
...
)

∫

tan x ln(cos x) dx

du

tan
xu


−
tan
x

∫
= − u du

=

1
= − ( ln (cos x))2 + C
2

Aside
Let u = ln (cos x)
− sin x
du =
dx
cos x
du = tan x dx
du
dx =
− tan x

Aside
x3 − 4x2 = x2 (x − 4)
1
Ax + B
C
=
+
x3 − 4x2
x2
x−4
1
Ax
B
C
= 2 + 2+
x3 − 4x2
x
x
x−4
A
B
C
= + 2+
x
x
x−4
When x = 4
1
C=
16
When x = 0
1
B=−
4
When x = 1
1
1
1
− =A− −
3
4 48
12
1
16
=A−
−
−
48
48 48
1
A=−
16

∫

1
dx
x3 − 4x2
∫
1
1
− 16
−1
=
+ 24 + 16 dx
x
x
x−4

(Q65
...
)

sin x cos(2x) dx
∫
= − (2 u2 − 1) du
2u3
=−(
− u)
3
2
= − cos 3 x + cos x + C
3

IIT-JEE Main Notes

Aside
cos (2x) = 2 cos2 x − 1
Let u = cos x
du = − sin x dx

22

Mahayodha Academy

∫
(Q67
...
)

=

√

Aside
2 cos (2x) = 1 + cos2 x

2 sin x + C

∫

1
dx
1 + tan x
∫
1
1 − tan x +1 + tan x
=
dx
2
1 + tan x
∫
1
1 − tan x 1 + tan x
=
+
dx
2
1 + tan x 1 + tan x
=

(Q69
...
)
dx
1
x
e
∫ 1 √
1 − u2
=

x du
−1

x
∫

=

e

y

1

π
2
−1

x

1

πr2
2
π
=
2

Area =

∫
(Q71
...
)

Aside
Let u = x + 1
du = dx

∫
(Q73
...
)

x−1
= x + 1) x2
−( x2 + x)

1
= x ln (1 + x) − x2 + x − ln (x + 1 ) + C
2
2

−x
−( −x − 1)
1

IIT-JEE Main Notes

25

Mahayodha Academy

∫
(Q75
...
)
∫ √
=
∫

Aside
Let u = ln x
1
du = dx
x
sec2 x = tan2 x + 1
Let w = tan u
dw = sec2 u du

1−x
dx
1+x
1 − x (1 − x)
dx
1 + x (1 − x)

1−x
√
dx
1 − x2
∫
∫
1
x
√
= √
dx −
dx
2
1−x
1 − x2
√
= sin−1 x + 1 − x2 + C
=

∫
(Q77
...
)

∫

√
sin−1 ( x) dx

∫

u2
du
1 − u2
Let u = sin θ
√

−

2u sin−1 u du
∫
u2
2
−1
√
=u sin u −
du
1 − u2
√
√
1
1√ √
= x sin−1 x − sin−1 x +
x 1−x+C
2
2
=

du = cos θ dθ
∫
sin 2 θ 

θ dθ
cos
−



θ
cos
∫
1
1 − cos 2θ dθ
=−
2
1
1
= − (θ − sin 2θ)
2
2
1
1
= − θ + sin θ cos θ
2
2
u
sin θ =
1
θ = sin−1 u
1

u

√θ
1 − u2

∫

tan−1 x dx
∫
−1
=x tan x −

(Q79
...
)
0

∫
=

∫

2

5

10 dx +
0

= 20 + [x3 −

if x ≤ 2
if x > 2

10
3x2 − 2

dx

( 3x2 − 2) dx

2
2x]52

= 20 + [53 − 10 − 23 + 4]
= 131

∫
(Q81
...
)
∫
=
=

x−1
dx
x4 − 1
1
2

1
dx −
2
x+1

∫

1
x
dx +
2
x2 + 1

∫

x4 − 1 = (x − 1)(x + 1)(x2 + 1)
Bx + C
A
1
+ 2
=
2
x +1
x+1
(x − 1)(x + 1)(x + 1)
1
A=
2
1
using x = 0, C =
2
1
using x = 1, B = −
2

1
dx
x2 + 1

1
1
1
ln |x + 1| − ln(x2 + 1) + tan−1 x + C
2
4
2

IIT-JEE Main Notes

28

Mahayodha Academy

∫
(Q83
...
)

√

∫

etan x
dx
1 − sin2 x
sec2 xetan x dx

=

Aside
1 − sin2 x = cos2 x

= etan x + C

∫

tan−1 x
dx
x2
∫
tan−1 x
1
=−
+
dx
x
x(1 + x2 )
∫
tan−1 x
1
=−
+
dx
2
x
x × x (x−2 + 1)
∫
tan−1 x
1
−2x−3
=−
−
dx
x
2
x−2 + 1

(Q85
...
)
=

∫

Aside
+
−

D
tan−1 x
1
1+x2

I
1
x2
− 1x

tan−1 x 1
− ln (x−2 + 1) + C
x
2

tan−1 x
dx
1 + x2
u du

1 2
u
2
1
= (tan−1 x)2 + C
2

=

IIT-JEE Main Notes

Aside
Let u = tan−1 x
1
dx
du =
1 + x2
x = tan u

29

Mahayodha Academy

Aside
∫
(ln x)2 dx
∫
=x(ln x)2 − 2 ln x dx
∫
=x(ln x)2 − ( 2x ln x − 2 dx)

(Q87
...
)
x2
∫
2 sec θ

2 sec2 θ dθ
=

4 tan2 θ

∫
2

θ
cos
1
×
=
2 dθ
3
sin θ
cos
θ
∫

2


cos
2 θ
sin θ
)dθ
+ 
= (

2

sin θ 
cos θ
θ sin2 θ
cos
= ln | sec θ + tan θ| − csc θ
√
√
x2 + 4
x
x2 + 4
+ C1
+ )−
= ln (
x
2
2
√
√
x2 + 4
+ C2
= ln ( x2 + 4 + x) −
x

D
+ (ln x)2
− 2 lnxx

I
1
x

I
1
x

D
+ 2 ln x
2
−
x

Aside
Let x = 2 tan θ
dx = 2 sec2 θ dθ
√
√
(2 tan θ)2 + 4 = 4(tan2 θ + 1)
√
= 4( sec2 θ)
= 2 sec θ
sin2 θ + cos2 θ = 1
x
tan θ =
2
√ x2

+

4

x

θ
2

∫ √

x+4
dx
x
∫
u
2u du
=
u2 − 4
∫
8
du
2+ 2
=
u −4
∫
8
du
2+
=
(u − 2)(u + 2)
∫
−2
2
du
+
2+
=
(u − 2) (u + 2)
=2u + 2 ln |u − 2| − 2 ln |u + 2|
√
√
x−4−2
|+C
= 2 x − 4 + 2 ln | √
x−4+2

(Q89
...
)

3

0

π
=
4

∫

x
dx
1 + x4
∫
2x
1
=
dx
2
1 + ( x2 )2

(Q91
...
)

∫

=
= 2

∫
(Q93
...
)

(Q95
...
)

∫

=

√
4

x dx

4 1
x4
5

+1

dx

4 5
= x4
5
4√
4
=
x5
5
4 √
= x4x+C
5

∫

1
dx
1 + ex
∫
1 +ex − ex
=
dx
1 + ex
∫
∫
ex
= 1dx −
dx
1 + ex

(Q97
...
)

∫

√

1 + ex dx
Aside

2u
=
u 2
du
u −1
∫
2
= 2+ 2
du
u −1
∫
2
= 2+
du
(u − 1)(u + 1)
∫
1
1
= 2+
−
du
(u − 1) (u + 1)
| u − 1|
= 2 u + ln
| u + 1|
√
√
1 + ex − 1
x
)+C
= 2 1 + e + ln( √
1 + ex + 1

√

1 + ex
ex
du = √
dx
2 1 + ex
√
2 1 + ex
dx =
du
ex
u2 − 1 = ex
Let u =

2
= u − 1) 2u2
2

−( 2u2 − 2)
2

∫ √

tan x
dx
sin(2x)
∫
u
dx
=
2 sin x cos x
∫
u
2u
=
× 2 du
sec x
2 sin x cos x

∫
2
cos
2 x
u
=
×
du


cos
x
sin x
1
∫
= u2 cot x du
∫
1
= u2 2 du
u
∫
= 1 du

(Q99
...
)

π

= [ tan x − sec x]02
[
]π
sin x
1 2
=
−
cos x cos x 0
[
]π
sin x − 1( sin x + 1) 2
=
cos x( sin x + 1) 0
[
] 2π
− cos 2 x
=


cos
x( 1 + sin x)

0
[
] 2π
− cos x
=
1 + sin x 0
=1

∫ (
(Q101
...
Gaikwad Nitin
+91-8788575873

IIT-JEE Main Notes

34

Mahayodha Academy



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