Integration

1. Find ∫ (x2 + 3x ) dx

∫Xn dx= xn+1/n+1 + c

x2+1 + c
2+1

= x3 + c
3

∫3x
= 3∫ x1+1
1+1

3∫ x2
2

= 3 ∫ x2
2
Threrfore ∫ (x2 + 3x ) dx = x3/3 + 3/2∫ x2 + c

2. Find ∫(2 x1/2) dx?

∫ 2x1/2 x dx

2∫ x1/2+1
½+1

2∫ x3/2
3/2

= 4 /3 x3/2 + c

3. Find ∫ (3x2 + 6x + 2) dx?

3∫x2+1 + 6∫x1+1 + 0
2+1 1+1

3∫x3 + 6∫ x2
3 2

∫x3 + 3∫x2
 ∫x3+1 + 3∫x2+1
3+1 2+1

= x4 + 3 x3 + c
4 3

= x4 + x3 + c
4

4. Find ∫ (x4 – 2x2) dx?

∫x4+1 - 2∫x2+1
4+1 2+1

x5 - 2 x3
5 3

= x5 - 2 x3 + c
5 3

Formulas

∫ cos x dx= sin x + c

∫ sin x dx = -cos x + c

∫ sec2 x dx = tan x + c

∫ cosec2 x dx = - cot x + c

∫ sec x tan x dx = sec x + c

∫ cosec x cot x dx = -cosec x + c

5. Find ∫ (1/ 1+ cos6x) dx?

∫ cos x dx = sin x +c
Let u = 6x
du= 6dx
cos(6x) dx = 1/6 ∫ cos(6x) 6dx
= 1/6∫ cos (u) du
= 1/6sin (u) + c
= 1/6 sin (6x) + c

cos (A+B) = cos A cos B – sin A sin B

A=B
cos (2B)= cos2 B – sin2 B

cos2 theta = 1- sin2 theta

sin2 + cos2 = 1

sin2 B = 1- cos2 theta

cos (2B)= cos2 B – sin2 B

cos 2B = cos2 B – (1- cos2 B)

cos 2B = cos2 B -1 + cos2 B

cos 2B = 2 cos2 B – 1

1 + cos 2B = 2 cos2 B

Let 2B = x
B=x/2

1 + cos x= 2 cos2 x
2

1
1+ cos6x

1+ cos6x = 2 cos2 6x
2

= 2 cos2 3x

∫ 1
1+ cos6x

=∫ 1
2 cos2 3x

= ∫ 1x1
2 cos2 3x

= ∫ 1 x sec2 3x
2

∫ sec2 x dx = tan x + c

= 1/2 ∫ sec2 u du/3

= ½ x 1/3∫ sec2 u du
 = tan (3x/6) + c

Let 3x = u

3dx = du

dx = du/3

6. Find ∫ ( 2 ) dx?
√x
= ∫ 2 dx
x1/2
= 2∫ x-1/2 dx

= 2 ∫ x-1/2 +1 dx
-1/2 +1

= 2 x½ + C
1/2

= 4 x1/2 + C

= 4√ x + C

7. Solve ∫ (x4 + 2x – 8 sin x) dx?

= ∫ x4 dx + ∫ 2x dx - ∫ 8 sin x dx

= ∫ x4+1 dx + 2 ∫ x1+1 dx - 8∫ sin x dx

4+1 1+1

= x5 + 2 x2 - 8 (-cos x) + C
5 2

= x5 + x2 + 8 cos x + C
5

8. Find ∫ (8√x) dx?

= 8 ∫ √x dx

= 8 ∫ x1/2 dx

= 8 ∫ x1/2 +1 dx
½+1
 = 8 x3/2 + C
3/2

= 16 (x3/2) + C
3

∫ ex dx= ex + C

∫ e-x dx = -e-x + C

∫ 1/ x dx = log x + C

∫ (a)x dx = (a)x + C
Log a

9. Find ∫ (cos x + ex)?

= ∫ cos x dx + ∫ex dx

= sin x + ex + C

10. Find ∫ (log x) dx?

= ∫ log x dx

= ∫ 1 x (x) (log x) dx
(x)

= x (log x) - ∫ (x) x (1/x) dx

= x (log x) – x + c

11. Find ∫ (ex √1+ex) dx?

= ∫ ex (1+ex)1/2 dx ……..equation 1.

Let (1+ex) = a……..equation 2.

da = ex dx……….equation 3.

Substuting equ 2and 3 in equ 1.

= ∫ (a)1/2 da
 = a1/2 + 1 + c
½+1

= a3/2 + c
3/2

= 2 a3/2 + C
3

= 2 (1+ex)3/2 + C
3

2 sin A cos B = sin (A + b) + sin (A - B)

2 cos A sin B = sin (A + B) – sin (A - B)

2 cos A cos B = cos (A +B) + cos (A- B)

2 Sin A sin B = cos (A - B) – cos (A + B)

12. Find ∫ sin 7 x . cos 2 x . dx?

Sol –

2 sin A cos B = sin (A + B) + sin (A - B)

=∫ sin (7x + 2x) + sin (7x – 2x) dx

2

=∫ sin 9 x + sin 5 x dx
2

= ∫ sin 9x dx + ∫ sin 5x dx
2 2

= ½ ∫ sin 9 x dx + ½ sin 5 x dx

∫ sin x dx = -cos x + c

∫ sin ax dx = 1/a (-cos ax)

= ½ (- cos 9x) + ½ (-cos 5x) + C

9 5
 = -1 (cos 9x) + (cos 5x) + C
2 9 5

13. Find ∫ sin 8x . cos 4x .dx?

Sol-
2 sin A cos B = sin (A + B) + sin (A - B)

= ∫ sin (8x + 4x) + sin (8x – 4x) dx

2

= ∫sin 12x + sin 4x dx

2

= ∫sin 12x dx + ∫sin 4x dx

2 2

= ½ ∫ sin 12x dx + ½ ∫sin 4x dx

= ½ (-cos 12x) + ½ (-cos 4x) + C

12 4
= -cos 12x - cos 4x + C
24 8

= -1 (cos12x) + (cos 4x) + C

8 3

14. Find ∫ cos 6x . cos 4x .dx?

2 cos A cos B = cos (A + B) + cos(A - B)

= ∫ cos (6x + 4x) + cos (6x - 4x) dx

2

= ∫ cos 10x + cos 2x dx

2

=∫cos 10x dx + cos 2x dx

2 2

= ½ ∫ sin 10x + ½ ∫ sin 2x + C

10 2

= sin 10x + sin 2x + C

20 4

= 1 (sin 10x) + (sin 2x ) + C

4 5
15. Find ∫ cos 6x . sin 9x . dx?

2 cos A sin B = sin (A + B) – sin (A – B)

=∫ sin (6x + 9x) – sin (6x – 9x) dx
2

=∫ sin 15x + sin 3x dx

2

=∫sin15x dx +∫ sin 3x dx
2 2

=1/2 ∫ sin 15 x dx + ½ ∫ sin 3x dx

= ½ (-cos 15x) + ½ (-cos 3x) + C

15 3

= -cox 15x – cos 3x + C

30 6

= -1 (cos 15x) + (cox 3x) + C

6 5

16. Find ∫ sin 8x . sin 6x . dx?

2 Sin A sin B = cos (A - B) – cos (A + B)

= ∫ cos (8x – 6x) – cos (8x + 6x) dx

2

=∫ cos 2x – cos 14x dx

2

=∫ cos 2x dx – cos14x dx
2 2

= ½ ∫ cos 2x dx – ½ ∫ cos 14x dx

= ½ (sin 2x) – ½ (sin 14x) + C

2 14

= sin 2x – sin 14x + C

4 28

= 1 (sin 2x) – (sin 14x) + C

4 7
 5-09-22

17. Find ∫ cos √x dx

√x

Let t= √x ….. equation1

Differeation both sides with respect to sides

d/dx (t)= d/dx (√x)

dt/dx= 1
2√x
dt= dx/2√x

2dt= dx / √x ……. Eqution2

Integrating the function
∫ cos √x dx
√x
Substituting √x = t and dx/√x = 2dt in the above function.

= ∫ cos t (2dt)

= 2∫ cos t dt

= 2 [sin t ] + C

= 2 [sin √x] + C

18. Find ∫ esin -1x dx

√1-x2

Let sin-1x= t ……equation 1

Differenction both sides of equation 1

d/dx (t)= d/dx (sin-1x)

dt/dx = 1
√1-x2

dt= dx
√1-x2 …….equation 2

Substituting equation 1 & 2 in given function

 ∫ esin -1x dx
√1-x2

= ∫ e t dt

= et + C

= e sin -1x + C

11. Find ʃ ( sec 2

x cosec 2 x ) dx cosec2 x = 1 + cot2 x

= ∫sec2 x (1+ cot 2 x) dx

= ∫ (sec2x + sec2x cot 2 x) dx

= ∫sec2 x dx + ∫ sec2 x 1/ tan2x dx

= let tan x = t

Sec2x dx = dt

dx= dt/ sec2 x

= tan x + ∫1 / t2 dt

= tan x + t-2 + 1 + C
-2 + 1

= tan x + t -1 + C
-1

= tan x – 1/ tan x + C

= Tan x – cot x + C
Find ʃ ( sec 2
x cosec 2 x ) dx

Sec2 x = 1/ cos2 x

Cosec2 x = 1/ sin2 x

=∫ 1 x 1 dx
cos2x sin2x

= ∫ sin2x + cos2x dx sin2 x+ cos 2 x = 1

cos2x sin2x

= ∫ sin2 x + cos2 x dx + ∫ sin2 x + cox2 x dx

Cos2 x sin2 x

= ∫ 1/ cos2 x dx + ∫ 1/sin2 x dx

= ∫sec2 dx + ∫ cosec2 dx

= tan x + (-cot x) + C

= tan x – cot x + C

2 . Find ʃ sin −1
( cosx ) dx

Cos theta = sin 90- theta = π/2

=∫ sin-1 [sin (π/2 – x)] dx
x-1 = 1/x

= ∫ [1/sin (sin (π/2 – x)] dx

= ∫(π/2 – x) dx

= ∫π/2 dx - ∫x dx
= π /2 ∫ 1 dx – ∫x dx

= π/2 x – x2 + C
2

sinx+cosx
3. Find ʃ √1+sin 2 x dx

sinx+cosx
Given = ʃ √1+sin 2 x dx

=∫ sin x + cos x dx
√sin2x + cos2x + 2 sinx cos x

=∫ sin x + cos x dx
√(Sin x +cos x)2

= ∫ sin x + cos x dx
Sin x + cos x

= ∫1 dx

=x+c

x2
4. Find ʃ √1+ x 2 ( 1+√ 1+ x2 ) dx

2
x
Given - ʃ √1+ x ( 1+√ 1+ x ) dx
2 2

Let x= tan θ

x/tan = θ
1/ tan = tan-1

Dx= sec2 θ dθ

Sec2 θ - tan2 θ = 1

Sec2 θ= 1+ tan2 θ = 1 +x2

Sec2 θ dθ= (1+x2) dθ

= ∫tan2 θ sec2 θ dθ
Sec2θ (1 + sec θ)

= ∫tan2 θ dθ
1+ sec θ

tan θ = sin θ /cos θ

sec θ= 1/cos θ

=∫ sin2 θ dθ
2
Cos θ (1 + 1/cos θ)

=∫ sin2 θ dθ
cos θ + 1
2
Cos θ ( )
cos θ

=∫ sin2 θ dθ
Cos θ (cos θ +1)

Sin2 θ +cos2 θ = 1

Sin2 θ = 1 - cos2 θ

= ∫ 1- cos2 θ d θ
cos θ (1 + cos θ)

= ∫ (1+ cos θ) (1-cos θ) dθ

Cos θ (1 + cos θ)

= ∫(1- cos θ) dθ
Cos θ

= ∫ 1/ cos θ dθ - ∫ cos θ /cos θ dθ

= ∫ sec θ dθ – ∫1 dθ
 ∫ sec θ = log (sec θ + tan θ)
= log ( sec θ + tan θ) - θ + C

= log (√1 +x2 + x ) – θ + C

= log (√1 + x2 + x)- tan-1 x + C

1. Evaluate ∫ ( 3 x +2 x+1 ) dx
1
2