RD Sharma Class 12 Ex 19.9 Solutions Chapter 19 Indefinite Integrals

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RD Sharma Class 12 Ex 19.9 Solutions Chapter 19 Indefinite Integrals

Evaluate the following integrals:

Question 1. ∫(log⁡x)/x dx

Solution:

Given that, I = ∫(log⁡x)/x dx

 Let us considered log⁡x = t

Now differentiating both side we get,

d(log⁡x) = dt

1/x dx = dt

dx = xdt

Then, put log⁡x = t and dx = xdt, we get

I = ∫ t/x × (x)dt

= ∫ tdt

= t²/2 + c

= (log⁡x)²/2 + c

Hence, I = (log⁡x)²/2 + c

Question 2.∫(log⁡(1 + 1/x))/(x(1 + x)) dx

Solution:

Given that I = ∫(log⁡(1 + 1/x))/(x(1 + x)) dx   ……(i)

Let us considered log⁡(1 + 1/x) = t then

On differentiating both side we get,

 d[log⁡(1 + 1/x)]=dt

1/(1 + 1/x) × (-1)/x²dx = dt

1/((x + 1)/x) × (-1)/x²dx = dt

(-x)/(x² (x + 1)) dx = -dt

dx/(x(x + 1)) = -dt

Now, on putting log⁡(1 + 1/x) = t and dx/(x(x + 1)) = -dt in equation (i), we get

I = ∫t × -dt

= -t²/2 + c

= -1/2 [log⁡(1 + 1/x)]² + c

Hence, I = -1/2 [log⁡(1 + 1/x)]² + c

Question 3. ∫((1 + √x )²)/√x dx

Solution:

Given that I = ∫((1 + √x)²)/√x dx

 Let us considered (1 + √x) = t then, 

On differentiating both side we get,

d(1 + √x) = dt

1/(2√x) dx = dt

dx = dt × 2√x

Now on putting (1 + √x) = t and dx = dt × 2√x, we get

I = ∫t²/√x × dt × 2√x

= 2∫t²dt

= 2 × t³/3 + c

= 2/3[1 + √x]³ + c

Hence, I = 2/3(1 + √x)³ + c

Question 4. ∫√(1 + e^x) e^x d

Solution:

Given that I = ∫√(1 + e^x) e^x dx  ……(i)

Let us considered 1 + e^x = t then, 

On differentiating both side we get,

d(1 + e^x) = dt

e^x dx = dt

dx = dt/e^x 

Now on putting 1 + e^x = t and dx = dt/e^x in equation (i), we get

I = ∫√t × e^x × dt/e^x

= ∫ t^1/2 dt

= 2/3t^3/2 + c

= 2/3 (1 + e^x)^3/2+c

Question 5. ∫∛(cos²x) sin⁡x dx

Solution:

Given that I = ∫∛(cos²x) sin⁡x dx    ……(i)

Let us considered cos⁡x = t then,

On differentiating both side we get,

d(cos⁡x) = dt

-sin⁡xdx = dt

dx = -dt/(sin⁡x))

Now on putting cos⁡x = t and dx = -dt/(sin⁡x) in equation (i), we get

I = ∫∛(t²) sin⁡x × (-dt)/(sin⁡x)

= -∫t^2/3 sin⁡x dt/(sin⁡x)

= -∫ t^2/3 dt

= -3/5 × t^5/3

Hence, I = -3/5(cos⁡x)^5/3 + c

Question 6. ∫e^x/(1 + e^x)² dx

Solution:

Given that I = ∫e^x/(1 + e^x)² dx   …….(i)

Let us considered 1 + e^x = t then,

On differentiating both side we get,

d(1 + e^x) = dt

e^x dx = dt

dx = dt/e^x 

Now on putting 1 + e^x = t and dx = dt/e^x in equation (i), we get

I = ∫e^x/t² × dt/e^x 

= ∫dt/t²

= ∫t^-2 dt

= -t^-1 + c

= -1/t + c

= -1/(1 + e^x) + c

 Hence, I = -1/(1 + e^x) + c

Question 7. ∫cot³⁡x cosec²⁡x dx

Solution:

Given that I = ∫cot³⁡x cosec²⁡x dx   …….(i)

Let us considered cot⁡x = t then,

On differentiating both side we get,

d(cotx) = dt

 -cosec²x dx = dt

dx = -dt/cosec²x

Now on putting cot⁡x = t and dx = -dt/(cosec²⁡x) in equation (i), we get

I = ∫ t³ cosec²x × (-dt)/(cosec²⁡x)

= -∫ t³ dt

= -t⁴/4 + c

= -(cot⁴⁡x)/4 + c

Hence, I = -(cot⁴⁡x)/4 + c

Question 8. 

Solution:

Given that I =[Tex] [/Tex] …….(i)

Let us considered sin^-1⁡x = t then,

On differentiating both side we get,

d(sin^-1x) = dt

1/√(1 – x²)dx = dt

dx = √(1 – x²) dt)

Now on putting sin^-1x = t and dx = √(1 – x²) dt in equation (i), we get

I = ∫ (e^t)²/√(1 – x²) × √(1 – x²) dt

= ∫e^2t dt

= e^2t/2 + c

=  + c

Hence, I =  + c

Question 9. ∫(1 + sin⁡x)/√(x – cos⁡x) dx

Solution:

Given that I = ∫(1 + sin⁡x)/√(x – cos⁡x) dx    ……..(i)

Let us considered x – cos⁡x = t, then 

On differentiating both side we get,

d(x – cos⁡x) = dt

 [1 – (-sin⁡x)]dx = dt

 (1 + sin⁡x)dx = dt

Now on putting x – cos⁡x = t and (1 + sin⁡x)dx = dt in equation (i), we get

I = ∫ dt/√t

= ∫ t^-1/2 dt

= 2t^1/2 + c

= 2(x – cos⁡x)^1/2 + c

Hence, I = 2√(x – cos⁡x) + c

Question 10.  ∫1/(√(1 – x²) (sin^-1x)²) dx

Solution:

Given that I = ∫1/(√(1 – x²) (sin^-1⁡x)²) dx   …..(i)

Let us considered sin^-1⁡x = t then,

On differentiating both side we get,

 d(sin^-1⁡x) = dt

1/√(1 – x² ) dx = dt

Now on putting sin^-1⁡x = t and 1/√(1 – x²) dx = dt in equation (i), we get

I = ∫dt/t²  

= ∫t^-2 dt

= -t^-1 + c

= (-1)/t + c

= (-1)/(sin^-1⁡x) + c

Hence, I = (-1)/(sin^-1⁡x) + c

Question 11. ∫(cot⁡x)/√(sin⁡x) dx

Solution:

Given that I = ∫(cot⁡x)/√(sin⁡x) dx    …….(i)

Let us considered sin⁡x = t then,

On differentiating both side we get,

d(sin⁡x) = dt

cos⁡xdx = dt

Now, I = ∫(cot⁡x)/√(sin⁡x) dx

 = ∫(cos⁡x)/(sin⁡x√(sin⁡x)) dx

 = ∫ cos⁡x/(sin⁡x)^3/2dx

 = ∫ cos⁡x/(sin⁡x)^3/2 dx   …….(ii)

Now on putting sin⁡x = t and cos⁡xdx = dt in equation (ii), we get

I = ∫ dt/t^3/2 

= ∫ t^-3/2 dt

= -2t^-1/2 + c

= -2/√t + c

= -2/√(sin⁡x) + c

I = -2/√sin⁡x + c

Question 12. ∫(tan⁡x)/√(cos⁡x) dx

Solution:

Given that I = ∫(tan⁡x)/√(cos⁡x) dx

I = ∫sin⁡x/cos⁡x√(cos⁡x) dx

= ∫ sin⁡x/(cos⁡x)^3/2 dx

= ∫sin⁡x/(cos⁡x)^3/2 dx   ……..(i)

Let us considered cos⁡x = t then,

On differentiating both side we get,

d(cos⁡x) = dt

-sin⁡xdx = dt

sin⁡xdx = -dt

Now on putting cos⁡x = t and sin⁡xdx = -dt in equation (i), we get

I = ∫(-dt)/t^-3/2 

= -∫t^-3/2 dt

= -[-2t^-1/2] + c

= 2/t^1/2 + c

= 2/√(cos⁡x) + c)

Hence, I = 2/√(cos⁡x) + c

Question 13. ∫cos³x/√(sin⁡x) dx

Solution:

Given that I = ∫cos³⁡x/√(sin⁡x) dx

= ∫(cos²xcos⁡x)/√(sin⁡x) dx

= ∫((1 – sin²⁡x)cos⁡x)/√(sin⁡x) dx

= ∫((1 – sin²x))/√(sin⁡x) cos⁡xdx ……(i)

Let us considered  sin⁡x = t then,

On differentiating both side we get,

d(sin⁡x) = dt

cos⁡xdx = dt

Now on putting sin⁡x = t and cos⁡xdx = dt in equation (i), we get

I = ∫(1 – t²)/√t dt

= ∫(t^-1/2-t² x t^-1/2)dt

=∫(t^-1/2 – t^3/2)dt

= 2t^1/2 – 2/5 t^5/2 + c

= 2(sin⁡x)^1/2 – 2/5(sin⁡x)^5/2 + c

Hence, I = 2√(sin⁡x) – 2/5(sin⁡x)^5/2 + c

Question 14. ∫(sin³x)/√(cos⁡x) dx

Solution:

Given that I = ∫(sin³⁡x)/√(cos⁡x) dx

= ∫(sin²⁡xsin⁡x)/√(cos⁡x) dx

=∫((1 – cos²⁡x))/√(cos⁡x) sin⁡xdx …….(i)

Let us considered cos⁡x = t then,

On differentiating both side we get,

d(cos⁡x) = dt

-sin⁡xdx = dt

sin⁡xdx = -dt

Now on putting cos⁡x = t and sin⁡xdx = -dt in equation (i), we get

I = ∫((1 – t²))/√t × -dt

= ∫(t² – 1)/√t dt

= ∫(t²/t^1/2 – 1/t^1/2)dx

= ∫(t^2-1/2 – t^-1/2)dt

= ∫(t^3/2 – t^1/2)dt

= 2/5 t^5/2 – 2t^1/2 + c

= 2/5 cos^5/2⁡x – 2cos^1/2⁡x + c

Hence, I = 2/5 cos^5/2x – 2√(cos⁡x) + c

Question 15. ∫1/(√(tan^-1x) (1 + x²)) dx

Solution:

Given that I = ∫1/(√(tan^-1x) (1 + x²)) dx  …..(i)

Let us considered tan^-1⁡x = t, then    

On differentiating both side we get,

d(tan^-1⁡x) = dt

1/(1 + x²) dx = dt

Now on putting tan^-1⁡x = t and 1/(1 + x²) dx = dt in equation (i), we get

I = ∫1/√t dt

= ∫t^-1/2 dt

= 2t^1/2 + c

= 2√tan^-1⁡x + c

Hence, I = 2√tan^-1⁡x + c

Question 16. ∫√(tan⁡x)/(sin⁡xcos⁡x) dx

Solution:

Given that I = ∫√(tan⁡x)/(sin⁡xcos⁡x) dx

= ∫(√(tan⁡x)×cos⁡x)/(sin⁡xcos⁡x×cos⁡x) dx

= ∫√(tan⁡x)/(tan⁡xcos²⁡x) dx

= ∫(sec²⁡xdx)/√(tan⁡x) dx

Let us considered tan⁡x = t, then

On differentiating both side we get,

sec²⁡xdx = dt

Now

I = ∫ dt/√t

= 2√t + c

Hence, I = 2√tan⁡x + c

Question 17. 1/x × (log⁡x)² dx

Solution:

Given that I = ∫1/x × (log⁡x)² dx   …..(i)

Let us considered log⁡x = t then,

On differentiating both side we get,

d(log⁡x) = dt

1/x dx = dt

Now on putting log⁡x = t and 1/x dx = dt in equation (i), we get

I = ∫t² dt

= t³/3 + c

= (log⁡x)³/3 + c

Hence, I = (log⁡x)³/3 + c

Question 18. ∫sin⁵x cos⁡x dx

Solution:

Given that I = ∫sin⁵⁡x cos⁡x dx  ……(i)

Let us considered sin⁡x = t then,

On differentiating both side we get,

d(sin⁡x) = dt

cos⁡xdx = dt

Now on putting sin⁡x = t and cos⁡xdx = dt in equation (i), we get

I = ∫ t⁵ dt

= t⁶/6 + c

= (sin⁶⁡x)/6 + c

Hencec, I = 1/6 (sin⁶⁡x) + c

Question 19. ∫tan^3/2⁡x sec²⁡x dx

Solution:

Given that I = ∫tan^3/2⁡xsec²⁡xdx  ……(i)

Let us considered tan⁡x = t then,

On differentiating both side we get,

d(tan⁡x) = dt

sec²⁡xdx = dt

Now on putting tan⁡x = t and sec²xdx = dt in equation (i), we get

I = ∫ t^3/2 dt

= 2/5 t^5/2 + c

= 2/5(tan⁡x)^5/2 + c

Hence, I = 2/5 tan^5/2⁡x + c

Question 20. ∫(x³)/(x² + 1)²dx

Solution:

Given that I = ∫(x³)/(x² + 1)²dx …….(i)

Let us considered 1 + x² = t then,

On differentiating both side we get,

d(1 + x²) = dt

2xdx = dt

xdx = dt/2

Now on putting 1 + x² = t and xdx = dt/2 in equation (i),we get 

I = ∫x²/t³ × dt/2

= 1/2∫(t – 1)/t³ dt         [1 + x² = t]

= 1/2∫[(t/t³ – 1/t³)dt]

= 1/2∫(t^-2 – t^-3)dt

= 1/2 [-1t^-1 – t^-2/(-2)] + c

= 1/2 [-1/t + 1/(2t²)] + c

= -1/2t + 1/(4t²) + c

= -1/2(1 + x²) + 1/(4(1 + x²)²) + c

= (-2(1 + x²) + 1)/(4(1 + x²)²) + c

= (-2 – 2x² + 1)/(4(1 + x²)²) + c

= (-2x² – 1)/(4(1 + x²)²) + c

= -(1 + 2x² )/(4(x² + 1)²) + c

Henec, I = -(1 + 2x²)/(4(x² + 1)²) + c

Question 21. ∫(4x + 2)√(x² + x + 1) dx

Solution:

Given that I = ∫(4x + 2)√(x² + x + 1) dx

Let us considered x² + x + 1 = t then,

On differentiating both side we get,

(2x + 1)dx = dt

Now,

I = ∫ (4x + 2)√(x² + x + 1) dx

= ∫2√t dt

= 2∫√t dt

= 2t^3/2/(3/2) + c

Hence, I = 4/3 (x² + x + 1)^3/2 + c

Question 22. ∫(4x + 3)/√(2x² + 3x + 1) dx

Solution:

Given that l = ∫(4x + 3)/√(2x² + 3x + 1) dx  ……(i)

Let us considered 2x² + 3x + 1 = t then,

On differentiating both side we get,

d(2x² + 3x + 1) = dt

(4x + 3)dx = dt

Now on putting 2x² + 3x + 1 = t and (4x + 3)dx = dt in equation (i), we get

I = ∫dt/√t

= ∫t^-1/2 dt

= 2t^1/2 + c

= 2√t + c

Hence, I = 2√(2x² + 3x + 1) + c

Question 23. ∫1/(1 + √x) dx

Solution:

Given that I = ∫1/(1 + √x) dx   …….(i)

Let us considered x = t² then,

On differentiating both side we get,

dx = d(t²)

dx = 2tdt

Now on putting x = t² and dx = 2tdt in equation (i), we get

I = ∫2t/(1 + √(t²)) dt

= ∫2t/(1 + t) dt

= 2∫t/(1 + t) dt

= 2∫(1 + t – 1)/(1 + t) dt

= 2⌋[(1 + t)/(1 + t) – 1/(1 + t)]dt

= 2∫dt – 2∫1/(1 + t) dt

= 2t – 2log⁡|(1 + t)| + c

= 2√x – 2log⁡|(1 + √x)| + c

Hence, I = 2√x – 2log⁡|(1 + √x)| + c

Question 24. 

Solution:

Given that I =  …….(i)

Let us considered cos²x = t then,

On differentiating both side we get,

d(cos²⁡x) = dt

-2cos⁡x sin⁡x dx = dt

-sin⁡2x dx = dt

sin⁡2x dx = -dt

Now on putting cos²⁡x = t and sin⁡2x dx = -dt in equation (i), we get

I = ∫e^t(-dt)

= -e^t + c

= – + c

Hence, I = – + c

Question 25. ∫(1 + cos⁡x)/((x + sin⁡x)³) dx
Solution:

Given that I = ∫(1 + cos⁡x)/((x + sin⁡x)³) dx    …..(i)
Let us considered x + sin⁡x = t then,
On differentiating both side we get,
 d(x + sin⁡x) = dt
(1 + cos⁡x)dx = dt
Now on putting x + sin⁡x = t and (1 + cos⁡x)dx = dt in equation (i), we get
I = ∫ dt/t³ 
= ∫ t^-3 dt
= t^-2/-2 + c
= -1/(2t²) + c
= (-1)/(2(x + sin⁡x)²) + c
Hence, I = (-1)/(2(x + sin⁡x)²) + c
Question 26. ∫(cos⁡x – sin⁡x)/(1 + sin⁡2x) dx
Solution:
Given that I = (cos⁡x – sin⁡x)/(1 + sin⁡2x) 
= (cos⁡x – sin⁡x)/((sin²⁡x + cos²⁡x) + 2sin⁡xcos⁡x)  [Because sin²⁡x + cos²⁡x = 1 and sin⁡2x = 2sin⁡xcos⁡x]
Let us considered sin⁡x + cos⁡x = t
On differentiating both side we get,
(cos⁡x – sin⁡x)dx = dt
Now, 
= ∫(cos⁡x – sin⁡x)/(1 + sin⁡2x) dx
= ∫(cos⁡x – sin⁡x)/((sin⁡x + cos⁡x)²) dx
= ∫dt/t²  
= ∫t^-2 dt
= -t^-1 + c
= -1/t + c
Hence, I = (-1)/(sin⁡x + cos⁡x) + c
Question 27. ∫(sin⁡2x)/(a + bcos⁡2x)² dx
Solution:
Given that I = ∫(sin⁡2x)/((a + bcos⁡2x)²) dx   ……(i)
Let us considered a + bcos⁡2x = t then,
On differentiating both side we get,
(a + bcos⁡2x) = dt
b(-2sin⁡2x)dx = dt
sin⁡2x dx = -dt/2b
Now on putting a + bcos⁡2x = t and sin⁡2xdx = -dt/2b in equation (i), we get
I = ∫1/t² × (-dt)/2b
= (-1)/2b ∫ t^-2 dt
= -1/2b (-1t^-1) + c
= 1/2bt + c
= 1/(2b(a + bcos⁡2x)) + c
Hence, I = 1/(2b(a + bcos⁡2x)) + c
Question 28. ∫(log⁡x²)/x dx
Solution:
Given that I = ∫(log⁡x²)/x dx  ……..(i)
Let us considered log⁡x = t then,
On differentiating both side we get,
 d(log⁡x) = dt
1/x dx = dt
dx/x = dt
Now, I = ∫(log⁡x²)/x dx
= ∫(2log⁡x)/x dx
= 2∫(log⁡x)/x dx …….(ii)
Now on putting log⁡x = t and dx/x = dt in equation (ii), we get
I = 2∫tdt
= (2t²)/2 + c
= t² + c
I = (log⁡x)² + c
Question 29. ∫(sin⁡x)/(1 + cos⁡x)² dx
Solution:
Given that I = ∫(sin⁡x)/((1 + cos⁡x)²) dx …..(i)
Let us considered 1 + cos⁡x = t then, 
On differentiating both side we get,
d(1 + cos⁡x) = dt
-sin⁡xdx = dt
sin⁡xdx = -dt
Now on putting 1 + cos⁡x = t and sin⁡dx = -dt in equation (i), we get
I = ∫(-dt)/t² 
= -∫t^-2dt
= -(-1t^-1) + c
= 1/t + c
= 1/(1 + cos⁡x) + c
Hence, I = 1/(1 + cos⁡x) + c
Question 30. ∫cot⁡x log⁡ sin⁡x dx
Solution:
Given that I = ∫cot⁡x log⁡ sin⁡x dx
Let us considered log⁡ sin⁡x = t
1/(sin⁡x).cos⁡xdx = dt
cot⁡x dx = dt
∫cot⁡x log⁡ sin⁡x dx = ∫tdt
= t²/2 + c
= 1/2(log⁡sin⁡x)² + c
Question 31. ∫sec⁡x.log⁡(sec⁡x + tan⁡x)dx
Solution:
Given that I = ∫sec⁡x.log⁡(sec⁡x + tan⁡x)dx  ……..(i)
Let us considered log⁡(sec⁡x + tan⁡x) = t then, 
On differentiating both side we get,
d[log⁡(sec⁡x + tan⁡x)] = dt
sec⁡x dx = dt    [Since, d/dx(log⁡(sec⁡x + tan⁡x)) = sec⁡x]
Now on putting log⁡(sec⁡x + tan⁡x) = t and sec⁡x dx = dt in equation (i), we get
I = ∫tdt
= t²/2 + c
= 1/2[log⁡(sec⁡x + tan⁡x)]² + c
Hence, I = 1/2[log⁡(sec⁡x + tan⁡x)]² + c
Question 32. ∫cosec⁡x log⁡(cosec⁡x – cot⁡x)dx
Solution:
Given that I = ∫cosec⁡x log⁡(cosec⁡x – cot⁡x)dx    ……(i)
Let us considered log⁡(cosec⁡x – cot⁡x) = t then,
On differentiating both side we get,
dx[log⁡(cosec⁡x – cot⁡x)] = dt
cosec⁡x dx = dt     [ Since, d/dx(log⁡(cosec⁡x – cot⁡x)) = cosec⁡x] 
Now on putting log⁡(cosec⁡x – cot⁡x) = t and cosec⁡xdx = dt in equation (i), we get
I = ∫tdt
= t²/2 + c
Hence, I = 1/2[log⁡(cosec⁡x – cot⁡x)]² + c
Question 33. ∫x³cos⁡x⁴ dx
Solution:
Given that I = ∫x³cos⁡x⁴ dx   …….(i)
Let us considered x⁴ = t then,
On differentiating both side we get,
dx(x⁴) = dt
4x³dx = dt
x³ = dt/4
Now on putting x⁴ = t and x³dx = dt/4 in equation (i), we get
I = ∫ cos⁡t dt/4
= 1/4sint + c
Hence, I = 1/4sinx⁴ + c
Question 34. ∫x³ sin⁡x⁴ dx
Solution:
Given that I = ∫x³sin⁡x⁴ dx   …..(i)
Let us considered x⁴ = t then,
On differentiating both side we get,
d(x⁴) = dt
4x³dx = dt
x³ = dt/4
Now on putting x⁴ = t and x³dx = dt/4 in equation (i), we get
I = ∫sin⁡t dt/4
= 1/4 ∫sin⁡t dt
= -1/4 cos⁡t + c
Hence, I = -1/4 cos⁡x⁴ + c
Question 35. ∫(xsin^-1x²)/√(1 – x⁴) dx
Solution:
Given that I = ∫(xsin^-1⁡x²)/√(1 – x⁴) dx  …….(i)
Let us considered sin^-1x² = t then,
On differentiating both side we get,
 d(sin^-1x²) = dt
2x × 1/√(1 – x⁴) dx = dt
x/√(1 – x⁴) dx = dt/2
Now on putting sin^-1x² = t and x/√(1 – x⁴) dx = dt/2 in equation (i), we get
I = ∫t dt/2
= 1/2 × t²/2 + c
= 1/4 (sin^-1x²)² + c
Hence, I = 1/4 (sin^-1x²)² + c
Question 36. ∫x³sin⁡(x⁴ + 1)dx
Solution:
Given that I = ∫x³ sin⁡(x⁴ + 1)dx ……..(i)
Let us considered x⁴ + 1 = t then,
On differentiating both side we get,
d(x⁴ + 1) = dt
x³ dx = dt/4
Now on putting x⁴ + 1 = t and x³dx = dt/4 in equation (i), we get
I = ∫ sin⁡t dt/4
= -1/4 cos⁡t + c
= -1/4 cos⁡(x⁴ + 1) + c
Hence, I = -1/4 cos⁡(x⁴ + 1) + c
Question 37. ∫(x + 1)e^x/(cos²⁡(xe^x) dx
Solution:
Given that I = ∫((x + 1)e^x)/(cos²⁡(xe^x)) dx   ……(i)
Let us considered xe^x = t then, 
On differentiating both side we get,
d(xe^x) = dt
(e^x + xe^x)dx = dt
(x + 1)e^xdx = dt
Now on putting xe^x = t and (x + 1)e^xdx = dt in equation (i), we get
I = ∫dt/(cos²⁡t)
= ∫ sec²tdt
= tan⁡t + c
= tan⁡(xe^x) + c
Hence, I = tan⁡(xe^x) + c
Question 38. 
Solution:
Given that I =    ……..(i)
Let us considered  = t then,
On differentiating both side we get,
d() = dt
3x²dx = dt
x² dx = dt/3
Now on putting  = t and x² dx = dt/3 in equation (i), we get
I = ∫cos⁡t dt/3
= (sin⁡t)/3 + c
Hence, I = sin⁡()/3 + c
Question 39. ∫2xsec³(x² + 3)tan⁡(x² + 3)dx
Solution:
Given that I = ∫2xsec³(x² + 3)tan⁡(x² + 3)dx    ………(i)
Let us considered sec⁡(x² + 3) = t then,
On differentiating both side we get,
d[sec⁡(x² + 3)] = dt
2xsec⁡(x² + 3)tan⁡(x² + 3)dx = dt
Now on putting sec⁡(x² + 3) = t and 2xsec⁡(x² + 3)tan⁡(x² + 3)dx = dt in equation (i), we get
I = ∫t² dt
= t³/3 + c
= 1/3 [sec⁡(x² + 3)]³ + c
Hence, I = 1/3 [sec⁡(x² + 3)]³ + c
Question 40. ∫(1 + 1/x)(x + log⁡x)² dx
Solution:
Given that I = ((x + 1)(x + log⁡x)²)/x
= ((x + 1)/x)(x + log⁡x)²
= (1 + 1/x)(x + log⁡x)²
Let us considered (x + log⁡x) = t
On differentiating both side we get,
(1 + 1/x)dx = dt 
Now,
I = ∫(1 + 1/x)(x + log⁡x)² dx
= ∫t² dt
= t³/3 + c
Hence, I = 1/3(x + log⁡x)³ + c
Question 41. ∫tan⁡x sec²⁡x√(1 – tan²x) dx
Solution:
Given that I = ∫tan⁡x sec²⁡x√(1 – tan²x) dx      ………(i)
Let us considered 1 – tan²⁡x = t then,
On differentiating both side we get,
d(1 – tan²x) = dt
-2tan⁡x sec²⁡x dx = dt
tan⁡x sec²⁡x dx = (-dt)/2
Now on putting 1 – tan²x = t and tan⁡x sec²⁡x dx = -dt/2 in equation (i), we get
I = ∫√t × (-dt)/2
=-1/2 ∫t^1/2 dt
=-1/2×t^3/2/(3/2) + c
=-1/3 t^3/2 + c
Hence, I = -1/3 [1 – tan²⁡x]^3/2 + c
Question 42.∫log⁡x (sin⁡(1 + (log⁡x)²)/x dx
Solution:
Given that I = ∫log⁡x (sin⁡(1 + (log⁡x)²)/x dx  ……..(i)
Now on putting 1 + (log⁡x)² = t and (log⁡x)/x dx = dt/2 in equation (i), we get
I = ∫sin⁡t × dt/2
= 1/2 ∫ sin⁡tdt
= -1/2 cos⁡t + c
= -1/2 cos⁡[1 + (log⁡x)²] + c
Hence, I = -1/2 cos⁡[1 + (log⁡x)²] + c
Question 43.∫ 1/x² × (cos²(1/x))dx
Solution:
Given that I = ∫ 1/x² × (cos²⁡(1/x))dx ……(i)
Let us considered 1/x = t then, 
On differentiating both side we get,
d(1/x) = dt
(-1)/x²dx = dt
1/x² dx = -dt
Now on putting 1/x = t and 1/x²dx = -dt in equation (i), we get
I = ∫cos²t(-dt)
= -∫cos²tdt
= -∫(cos2t + 1)/2 dt
= -1/2 ∫cos⁡2t dt – 1/2 ∫dt
= -1/2 × (sin⁡2t)/2 – 1/2 t + c
= -1/4 sin⁡2t – 1/2 t + c
= -1/4 sin⁡2 × 1/x – 1/2 × 1/x + c
Hence, I = -1/4 sin⁡(2/x) – 1/2 (1/x) + c
Question 44. ∫sec⁴x tan⁡x dx
Solution:
Given that I = ∫sec⁴x tan⁡x dx  ……(i)
Let us considered tan⁡x = t then,
On differentiating both side we get,
d (tan⁡x) = dt
sec²xdx = dt
dx = dt/sec²⁡x
Now on putting tan⁡x = t and dx = dt/(sec²⁡x) in equation (i), we get
I = ∫sec⁴x tan⁡x dt/(sec²⁡x)
= ∫ sec²x tdt  
= ∫ (1 + tan²⁡x)tdt
= ∫(1 + t²)tdt
= ∫(t + t³)dt
= t²/2 + t⁴/4 + c
= (tan²⁡x)/2 + (tan⁴⁡x)/4 + c
Hence, I = 1/2 tan²⁡x + 1/4 tan⁴⁡x + c
Question 45. ∫(e^√x cos⁡(e^√x ))/√x dx
Solution:
Given that I = ∫(e^√x cos⁡(e^√x))/√x dx  …….(i)
Let us considered e^√x = t then, 
On differentiating both side we get,
d(e^√x) = dt 
e^√x(1/(2√x))dx = dt
e^√x/√x dx = 2dt
Now on putting e^√x = t and e^√x/√x dx = 2dt in equation (i), we get
I = ∫ cos⁡t × 2dt
= 2∫ cos⁡tdt
= 2sin⁡t + c
= 2sin⁡(e^√x) + c
I = 2sin⁡(e^√x) + c
Question 46. ∫(cos⁵x)/(sin⁡x) dx
Solution:
Given that I = ∫(cos⁵⁡x)/(sin⁡x) dx    …..(i)
Let us considered sin⁡x = t then,
On differentiating both side we get,
d(sin⁡x) = dt
cos⁡x dx = dt
dx = dt/(cos⁡x)
Now on putting sin⁡x = t and dx = dt/(cos⁡x) in equation (i), we get
I = ∫(cos⁵⁡x)/t × dt/(cos⁡x)
= ∫(cos⁴⁡x)/t dt
= ∫(1 – sin²⁡x)²/t dt
= ∫(1 – t²)²/t dt
= ∫(1 + t⁴ – 2t²)/t dt
= ∫1/t dt + ∫t⁴/t dt – 2∫t²/t dt
= log⁡|t| + t⁴/4 – (2t²)/2 + c
= log⁡|sin⁡x| + (sin⁴⁡x)/4 – sin²⁡x + c
Hence, I = 1/4 sin⁴x – sin²x + log⁡|sin⁡x| + c
Question 47. ∫(sin⁡√x)/√x dx
Solution:
Given that I = ∫(sin⁡√x)/√x dx
Let us considered √x = t then,
On differentiating both side we get,
1/(2√x) dx = dt
1/√x dx = 2dt
Now, 
I = ∫(sin⁡√x)/√x dx
= 2 ∫sint dt
= -2 cos⁡t + c
Hence, I = -2cos⁡√x + c
Question 48. ∫((x + 1)e^x)/(sin²(xe^x)) dx
Solution:
Given that I = ∫((x + 1)e^x)/(sin²⁡(xe^x)) dx   …….(i)
Let us considered xe^x = t then,
d(xe^x) = dt
(xe^x + e^x)dx = dt
(x + 1)e^xdx = dt
Now on putting xe^x = t and (x + 1)e^x dx = dt in equation (i), we get
I = ∫dt/(sin²t)
= ∫cos⁡ec²t dt
= -cot + c
Hence, I = -cot⁡(xe^x) + c

Question 49. 

Solution:

Given that I =   ……..(i)

Let us considered x + tan^-1⁡x = t then,

On differentiating both side we get,

d(x + tan^-1⁡x) = dt

(1 + 1/(1 + x²))dx = dt

((1 + x² + 1)/(1 + x²))dx = dt

((x² + 2))/((x² + 1)) dx = dt

Now on putting x + tan^-1⁡x = t and ((x² + 2)/(x² + 1))dx = dt in equation (i), we get

I = ∫5^tdt

= 5^t/(log⁡5) + c

= /(log⁡5) + c

Hence, I = /(log⁡5) + c

Question 50. 

Solution:

Given that I =   ……(i)

Let us considered msin^-1⁡x = t then, 

On differentiating both side we get,

d(msin^-1x) = dt

m 1/√(1 – x²) dx = dt

dx/√(1 – x²) = dt/m

Now on putting msin^-1⁡x = t and dx/√(1 – x²) = dt/m in equation (i), we get

I = ∫e^tdt/m

= 1/m e^t+c

= 

Hence, I = 

Question 51. ∫(cos⁡√x)/√x dx

Solution:

Given that I = ∫(cos⁡√x)/√x dx

Let us considered √x = t then, 

On differentiating both side we get,

1/(2√x) dx = dt

Now,

= ∫ (cos⁡√x)/√x dx

= 2∫ cos⁡tdt2

= 2sin⁡t + c 

Hence, I = 2sin⁡√x + c

Question 52. ∫sin(tan^-1x)/(1 + x²) dx

Solution:

Given that I = ∫sin(tan^-1x)/(1 + x²) dx   ……(i)

Let us considered tan^-1 = t then,

On differentiating both side we get,

 d(tan^-1⁡x) = dt

1/(1 + x²) dx = dt

Now on putting tan^-1x = t and dx/(1 + x²) = dt in equation (i), we get

I = ∫ sin⁡tdt

= -cos⁡t + c

= -cos⁡(tan^-1x) + c

Hence, I = -cos⁡(tan^-1⁡x) + c

Question 53. ∫(sin⁡(log⁡x))/x dx

Solution:

Given that I = ∫(sin⁡(log⁡x))/x dx   ……..(i)

Let us considered log⁡x = t then,

On differentiating both side we get,

d(log⁡x) = dt

1/x dx = dt

Now on putting log⁡x = t and 1/x dx = dt in equation (i), we get

I = ∫sin⁡tdt

= -cos⁡t + c

= -cos⁡(log⁡x) + c

Hence, I = -cos⁡(log⁡x) + c

Question 54.

Solution:

Given that I =

Let us considered tan^-1⁡x = t, then

On differentiating the above function we have, 

1/(1 + x²) dx = dt

 = ∫e^mt × dt

== e^mt/m

On Substituting the value of t, we 

I =  + c

Question 55. ∫x/(√(x² + a²) + √(x² – a²)) dx

Solution:

Given that I = ∫x/(√(x² + a²) + √(x² – a²)) dx

= ∫ x/(√(x² + a²) + √(x² – a²)) × (√(x² + a²) – √(x² – a² ))/(√(x² + a²) – √(x² – a²)) dx

= ∫ x(√(x² + a²) – √(x² – a²))/(x² + a² – x² + a²) dx

= ∫ x/(2a²) (√(x² + a²) – √(x² – a²))dx

I = 1/(2a²) ∫x(√(x² + a²) – √(x² – a²))dx ……(i)

Let us considered x² = t then, 

On differentiating the above function we have, 

d(x²) = dt

2xdx = dt

xdx = dt/2

Now on putting x² = t and xdx = dt/2 in equation (i), we get

I = 1/(2a²) ∫(√(t + a²) – √(t – a²)) dx/2

Hence, I = 1/(4a²) [2/3 (t + a²)^3/2 – 2/3 (t – a²)^3/2] + c

Question 56. ∫x(tan^-1⁡x²)/(1 + x⁴) dx

Solution:

Given that I = ∫x(tan^-1⁡x²)/(1 + x⁴) dx  ……..(i)

Let us considered tan^-1x² = t then,

On differentiating the above function we have, 

d(tan^-1x²) = dt

(1 × 2x)/(1 + (x²)²) dx = dt

(1 × x)/(1 + x⁴) dx = dt/2

Now on putting tan^-1⁡x² = t and x/(1 + x⁴) dx = dt/2 in equation (i), we get

I = ∫ t dx/2

= 1/2 ∫tdt

= 1/2 × t²/2 + c

= t²/4 + c – 1

= (tan^-1x²)²/4 + c

Hence, I = 1/4 (tan^-1⁡x²)² + c

Question 57. ∫(sin^-1x)³/√(1 – x²) dx

Solution:

Given that I = ∫(sin^-1x)³/√(1 – x²) dx ……(i)

Let us considered sin^-1x = t then,

On differentiating the above function we have, 

d(sin^-1⁡x) = dt

1/√(1 – x²) dx = dt

Now on putting sin^-1⁡x = t and 1/√(1 – x²) dx = dt in equation (i), we get

I = ∫t³ dt

= t⁴/4 + c

Hence, I = 1/4 (sin^-1x)⁴ + c

Question 58.∫(sin⁡(2 + 3log⁡x))/x dx

Solution:

Given that I = ∫(sin⁡(2 + 3log⁡x))/x dx  ……..(i)

Let us considered 2 + 3log⁡x = t then,

On differentiating the above function we have, 

d(2 + 3log⁡x) = dt

3 1/x dx = dt

dx/x = dt/3

Now on putting 2 + 3log⁡x = t and dx/x = dt/3 in equation (i), we get

I = ∫sin⁡t dt/3

= 1/3(-cos⁡t) + ct 

= -1/3 cos⁡(2 + 3log⁡x) + c

Hence, I = -1/3 cos⁡(2 + 3log⁡x) + c

Question 59. 

Solution:

Given that I =    ……(i)

Let us considered x² = t then,

On differentiating the above function we have, 

d(x²) = dt

2xdx = dt

 xdx = dt/2

Now on putting x² = t and xdx = dt/2 in equation (i), we get

I = ∫e^tdt/2

= 1/2 e^t+c

= 1/2  + c

Hence, I = 1/2  + c

Question 60. ∫e^2x/(1 + e^x) dx

Solution:

Given that I = ∫e^2x/(1 + e^x) dx  …….(i)

Let us considered 1 + e^x = t then, 

On differentiating the above function we have, 

d(1 + e^x) = dt

e^xdx = dt

dx = dt/e^x

Now on putting 1 + e^x = t and dx = dt/e^x in equation (i), we get

I = ∫e^2x/t × dt/e^x 

= ∫e^x/t dt

= ∫ (t – 1)/t dt

= ∫ (t/t – 1/t)dt

= t – log⁡|t| + c

= (1 + e^x) – log⁡|1 + e^x| + c

Hence, I = 1 + e^x – log⁡|1 + e^x| + c

Question 61. ∫(sec²⁡√x)/√x dx

Solution:

Given that I = ∫(sec²√x)/√x dx   ……(i)

Let us considered √x = t then,

On differentiating the above function we have, 

d(√x) = dt

1/(2√x) dx = dt

dx = 2√x dt

dx = 2tdt            [√x = t])

Now on putting √x = t and dx = 2tdt in equation (i), we get

I = ∫ (sec²t)/t × 2tdt

= 2∫ sec²⁡tdt

= 2tan⁡t + c

= 2tan⁡√x + c

Hence, I = 2tan⁡√x + c

Question 62. ∫tan³2x sec⁡2x dx

Solution:

Given that I = ∫tan³2x sec⁡2x dx

= tan²2xtan⁡2x sec⁡2x

= (sec²2x – 1)tan⁡2x sec⁡2x

= sec²2x.tan⁡2xsec⁡2x – tan⁡2xsec⁡2x

= ∫ sec²⁡2xtan⁡2xsec⁡2xdx – ∫tan⁡2xsec⁡2xdx

= ∫ sec²2xtan⁡2xsec⁡2xdx – (sec⁡2x)/2 + c

Let us considered sec⁡2x = t

2sec⁡2xtan⁡2xdx = dt

I = 1/2 ∫t² dt – (sec⁡2x)/2 + c

I = t³/6 – (sec⁡2x)/2 + c

Hence, I = (sec⁡2x)³/6 – (sec⁡2x)/2 + c

Question 63. ∫(x + √(x + 1))/(x + 2) dx

Solution:

Given that I = ∫(x+√(x+1))/(x+2) dx  …….(i)

Let us considered x + 1 = t² then,

On differentiating the above function we have, 

d(x + 1) = d(t²)

dx = 2tdt

Now on putting x + 1 = t² and dx = 2tdt in equation (i), we get

I = ∫ (x + √(t²))/(x + 2) 2tdt

= 2∫((t² – 1) + t)/((t² – 1) + 2) × tdt   [x + 1 = t²]

= 2∫(t² + t – 1)/(t² + 1) tdt

= 2∫ (t³ + t² – t)/(t² + 1) dt

= 2[∫ t³/(t² + 1) dt + ∫ t²/(t² + 1) dt – ∫ t/(t² + 1) dt]

I = 2[∫t³/(t² + 1) dt + ∫t²/(t² + 1) dt – ∫t/(t² + 1) dt]    ……(ii)

Let I₁ = ∫t³/(t² + 1) dt

I₂ = ∫t²/(t² + 1) dt

and I₃ = ∫t/(t² + 1) dt

Now, I₁ = ∫t³/(t² + 1) dt

= ∫(t – t/(t² + 1))dt

= t²/2 – 1/2 log⁡(t² + 1)      

I₁ = t²/2 – 1/2 log⁡(t² + 1) + c₁    ……..(iii)

Since, I₂ = ∫t²/(t² + 1) dt

 = ∫ (t² + 1 – 1)/(t² + 1) dt

 = ∫(t² + 1)/(t² + 1) dt – ∫1/(t² + 1) dt

 = ∫dt – ∫1/(t² + 1) dt

I₂ = t – tan^-1⁡(t²) + c₂ ………….(iv)

and, 

I₃ = ∫t/(t² + 1) dt

= 1/2 log⁡(1 + t²) + c₃   ……..(v)

Using equations (ii), (iii), (iv) and (v), we get

I = 2[t²/2 – 1/2 log⁡(t² + 1) + c₁ + t-tan^-1⁡(t²) + c₂ – 1/2 log⁡(1 + t²) + c₃]

= 2[t²/2 + t-tan^-1⁡(t²) – log⁡(1 + t²) + c₁ + c₂ + c₃]

= 2[t²/2 + t – tan^-1⁡(t²) – log⁡(1 + t²) + c₄  [Putting c₁ + c₂ + c₃ = c₄]

= t² + 2t – 2tan^-1(t²) – 2log⁡(1 + t²) + 2c₄

= (x + 1) + 2√(x + 1) – 2tan^-1(√(x + 1)) – 2log⁡(1 + x + 1) + 2c₄

= (x + 1) + 2√(x + 1) – 2tan^-1⁡(√(x + 1)) – 2log⁡(x + 2) + c [Putting 2c₄ = c]

Hence, I = (x + 1) + 2√(x + 1) – 2tan^-1⁡(√(x + 1)) – 2log⁡(x + 2) + c

Question 64. 

Solution:

Given that I =   …….(1)

Let us considered  = t then 

On differentiating the above function we have, 

 d() = dt

 ×  × 5^x × (log⁡5)³ dx = dt

dx = dt/((log⁡5)^3 ))

Now on putting  = t and dx = dt/((log⁡5)^3 )) in equation (i), we get

I = ∫dt/((log⁡5)³)

= 1/((log⁡5)³) ∫dt

= t/((log⁡5)^3) + c

Hence, I = /((log⁡5)³) + c)

Question 65. ∫1/(x√(x⁴ – 1)) dx

Solution:

Given that I = ∫1/(x√(x⁴ – 1)) dx   ……….(i)

Let us considered x² = t then,

On differentiating the above function we have, 

d(x²) = dt

2xdx = dt

dx = dt/2x

Now on putting x² = t and dx = dt/2x in equation (i), we get

I = ∫ 1/(x√(t² – 1)) × dt/2x

= 1/2 ∫ 1/(x² √(t² – 1)) dt

= 1/2 ∫ 1/(t√(t² – 1)) dt

= 1/2 sec^-1t + c

= 1/2 sec^-1x² + c

Hence, I = 1/2 sec^-1x² + c

Question 66. ∫√(e^x – 1) dx

Solution:

Given that I = ∫√(e^x – 1) dx   ……..(i)

Let us considered e^x – 1 = t² then, 

On differentiating the above function we have, 

d(e^x – 1) = dt(t²)

e^x dx = 2tdt

dx = 2t/e^x dt

dx = 2t/(t² + 1) dt               [e^x – 1 = t²] 

Now on putting e^x – 1 = t² and dx = 2tdt/(t² + 1) in equation (i), we get

I = ∫√(t²) × 2tdt/(t² + 1)

= 2∫(t × t)/(t² + 1) dt

= 2∫t²/(t² + 1) dt

= 2∫(t² + 1 – 1)/(t² + 1) dt

= 2∫[(t² + 1)/(t² + 1) – 1/(t² + 1)]dt

= 2∫dt – 2∫1/(t² + 1) dt

= 2t – 2tan^-1⁡(t) + c

= 2√(e^x – 1) – 2tan^-1(√(e^x – 1)) + c

Hence, I = 2√(e^x – 1) – 2tan^-1⁡√(e^x – 1) + c

Question 67. ∫ 1/(x + 1)(x² + 2x + 2) dx

Solution:

Given that I = ∫1/(x + 1)(x² + 2x + 2) dx

= ∫1/(x + 1)((x + 1)² + 1) dx

Let us considered x + 1 = tan u then,                               [tanu = Perpendicular/Base = (x + 1)/1]

On differentiating the above function we have, 

dx = sec²u du                                                                   [Hypotenuses = √(x² + 2x + 2)]

I = ∫sec²u/tanu(tan²u + 1) du

= ∫ cosu/sinu du

= log| sinu |+c

= log| sin(x + 1)| + c                                         [As we know, sin(x + 1) = P/H = (x + 1)/√(x² + 2x + 2)]

Hence,  I = log| x + 1/√(x² + 2x + 2)| + c

Question 68. ∫x⁵/√(1 + x³) dx

Solution:

Given that I = ∫ x⁵/(√(1 + x³)) dx   …….(i)

Let us considered 1 + x³ = t², then

On differentiating the above function we have, 

d(1 + x³) = d(t²)

3x² dx = 2t * dt

dx = 2t dt/3x²

Now on putting 1 + x³ = t² and dx = 2tdt/3x² in equation (i), we get

I = ∫ x⁵/√t² * 2t/3x² dt

= ∫x⁵/t * 2t/3x² dt

= 2/3∫x³ dt

= 2/3 ∫( t² – 1) dt

= 2/3[t³/3 – 2t/3] + c

Hence, I = 2/9(1 + x³)^3/2 – 2 √(1 + x²)/3 + c

Question 69. ∫4x³ √(5 – x²) dx

Solution:

Given that I = ∫4x³ √(5 – x²) dx  ……(i)

Let us considered 5 – x² = t² then, 

On differentiating the above function we have, 

d(5 – x²) = t²

-2xdx = 2tdt

dx = (-t)/x dt

Now on putting 5 – x² = t² and dx = (-t)/x dt in equation (i), we get

I = ∫4x³ √(t²) × (-t)/x dt

= -4∫ x² t × tdt

= -4∫(5 – t²) t² dt         [5 – x² = t²]

= -4∫(5t² – t⁵)dt

= -20×t³/3 + 4 t⁵/5 + c

= (-20)/3 × t³ + 4/5 × t⁵ + c

= (-20)/3 × (5 – x²)^3/2 + 4/5 × (5 – x²)^5/2 + c

I = (-20)/3 × (5 -x²)^3/2 + 4/5 × (5 – x²)^5/2 + c

Question 70. ∫1/(√x + x) dx

Solution:

Given that I = ∫1/(√x + x) dx  ……..(i)

Let us considered √x = t then,

On differentiating the above function we have, 

 d(√x) = dt

1/(2√x) dx = dt

dx = 2√x dt

Now on putting √x = t and 2√x dt = dx in equation (i), we get

I = ∫1/(t + t²) 2t × dt  [Since √x = t and x = t²]

= ∫2t/(t(1 + t)) dt

= 2∫t/(1 + t) dt

= 2log⁡|1 + t| + c

= 2log⁡|1 + √x| + c

Hence, I = 2log⁡|1 + √x|+c

Question 71. ∫1/(x² (x⁴ + 1)^3/4) dx

Solution:

Given that I = 1/(x² (x⁴+1)^3/4)

Multiplying and dividing by x^-3, we obtain

(x^-3/(x²x^-3 (x⁴+ 1)^3/4) = (x^-3 (x⁴ + 1)^-3/4/(x²x^-3))

= (x⁴ + 1)^-3/4/(x⁵(x⁴)^-3/4

= 1/x⁵ ((x⁴ + 1)/x⁴)^-3/4

= 1/x⁵ (1 + 1/x⁴)^-3/4

Let us considered 1/x⁴ = t

-4/x⁵ dx = dt

1/x⁵ dx = -dt/4

I = ∫ 1/x⁵(1 + 1/x⁴)^-3/4 dx

= -1/4 ∫(1 + t)^-3/4 dt

= -1/4 [(1 + t)^1/4)/(1/4)] + c

= -1/4(1 + 1/x⁴)^1/4/(1/4) + c

Hence, I = -(1 + 1/x⁴)^1/4 + c

Question 72. ∫(sin⁡⁵x)/(cos⁴x) dx

Solution:

Given that I = ∫(sin⁡⁵x)/(cos⁴⁡x) dx  ……(i)

Let us considered cos⁡x = t then, 

On differentiating the above function we have, 

d(cos⁡x) = dt

-sin⁡xdx = dt

 dx = -dt/(sin⁡x)

Now on putting  cos⁡x = t and dx = -dt/(sin⁡x) in equation (i), we get

I = ∫(sin⁵⁡x)/t⁴ × -dt/(sin⁡x)

= -∫(sin⁴⁡x)/t⁴ dt

= -∫(1 – cos²x)²/t⁴ dt

= -∫(1 – t²)²/t⁴ dt

= -∫(1 + t⁴ – 2t²)/t⁴dt

= -∫(1/t⁴ + t⁴/t⁴ – (2t²)/t⁴)dt

= -∫(t^-4 + 1 – 2t^-2)dt

= -[t^-3/(-3) + t – 2 t^-1/(-1)] + c

= 1/3 * 1/t³– t – 1/t + c

Hence, I = 1/3 * 1/cos³x – cosx -2/cosx + c

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