RD Sharma Class 12 Ex 19.9 Solutions Chapter 19 Indefinite Integrals

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.9
CategoryRD Sharma Solutions

Table of Contents

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)]+ c

Hence, I = -1/2 [log⁡(1 + 1/x)]+ 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]+ c

Hence, I = 2/3(1 + √x)+ c

Question 4. ∫√(1 + ex) ex d

Solution:

Given that I = ∫√(1 + ex) edx  ……(i)

Let us considered 1 + e= t then, 

On differentiating both side we get,

d(1 + ex) = dt

ex dx = dt

dx = dt/ex 

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

I = ∫√t × ex × dt/ex

= ∫ t1/2 dt

= 2/3t3/2 + c

= 2/3 (1 + ex)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)

= -∫t2/3 sin⁡x dt/(sin⁡x)

= -∫ t2/3 dt

= -3/5 × t5/3

Hence, I = -3/5(cos⁡x)5/3 + c

Question 6. ∫ex/(1 + ex)2 dx

Solution:

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

Let us considered 1 + e= t then,

On differentiating both side we get,

d(1 + ex) = dt

ex dx = dt

dx = dt/ex 

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

I = ∫ex/t2 × dt/ex 

= ∫dt/t2

= ∫t-2 dt

= -t-1 + c

= -1/t + c

= -1/(1 + ex) + c

 Hence, I = -1/(1 + ex) + 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 = ∫ tcosec2x × (-dt)/(cosec2⁡x)

= -∫ t3 dt

= -t4/4 + c

= -(cot4⁡x)/4 + c

Hence, I = -(cot4⁡x)/4 + c

Question 8. ∫\frac{(e^{sin^{-1} x})^2}{\sqrt{1-x^2}} dx

Solution:

Given that I =∫\frac{(e^{sin^{-1} x})^2}{\sqrt{1-x^2}} dx[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 = ∫ (et)2/√(1 – x2) × √(1 – x2) dt

= ∫e2t dt

= e2t/2 + c

\frac{e^{sin^{-1}x}}{2} + c

Hence, I = \frac{e^{sin^{-1}x}}{2} + 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

= 2t1/2 + 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/t3/2 

= ∫ 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/t1/2 + 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 – t3/2)dt

= 2t1/2 – 2/5 t5/2 + 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

= ∫(t– 1)/√t dt

= ∫(t2/t1/2 – 1/t1/2)dx

= ∫(t2-1/2 – t-1/2)dt

= ∫(t3/2 – t1/2)dt

= 2/5 t5/2 – 2t1/2 + c

= 2/5 cos5/2⁡x – 2cos1/2⁡x + c

Hence, I = 2/5 cos5/2x – 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

= 2t1/2 + 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. ∫tan3/2⁡x sec2⁡x dx

Solution:

Given that I = ∫tan3/2⁡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 = ∫ t3/2 dt

= 2/5 t5/2 + c

= 2/5(tan⁡x)5/2 + c

Hence, I = 2/5 tan5/2⁡x + c

Question 20. ∫(x3)/(x+ 1)2dx

Solution:

Given that I = ∫(x3)/(x+ 1)2dx …….(i)

Let us considered 1 + x= t then,

On differentiating both side we get,

d(1 + x2) = dt

2xdx = dt

xdx = dt/2

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

I = ∫x2/t3 × dt/2

= 1/2∫(t – 1)/t3 dt         [1 + x= 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 – 2x+ 1)/(4(1 + x2)2) + c

= (-2x– 1)/(4(1 + x2)2) + c

= -(1 + 2x2 )/(4(x+ 1)2) + c

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

Question 21. ∫(4x + 2)√(x+ x + 1) dx

Solution:

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

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

On differentiating both side we get,

(2x + 1)dx = dt

Now,

I = ∫ (4x + 2)√(x+ x + 1) dx

= ∫2√t dt

= 2∫√t dt

= 2t3/2/(3/2) + c

Hence, I = 4/3 (x+ x + 1)3/2 + c

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

Solution:

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

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

On differentiating both side we get,

d(2x+ 3x + 1) = dt

(4x + 3)dx = dt

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

I = ∫dt/√t

= ∫t-1/2 dt

= 2t1/2 + c

= 2√t + c

Hence, I = 2√(2x+ 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 = tand 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. ∫e^{cos^{2}x} sin⁡2xdx

Solution:

Given that I = ∫e^{cos^{2}x} sin⁡2xdx …….(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 = ∫et(-dt)

= -e+ c

= –e^{cos^{2}x} + c

Hence, I = –e^{cos^{2}x} + 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)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/t× (-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
= t+ c
I = (log⁡x)+ 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)+ 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)]+ c
Hence, I = 1/2[log⁡(sec⁡x + tan⁡x)]+ 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)]+ c
Question 33. ∫x3cos⁡x4 dx
Solution:
Given that I = ∫x3cos⁡x4 dx   …….(i)
Let us considered x= t then,
On differentiating both side we get,
dx(x4) = dt
4x3dx = dt
x= dt/4
Now on putting x= t and x3dx = dt/4 in equation (i), we get
I = ∫ cos⁡t dt/4
= 1/4sint + c
Hence, I = 1/4sinx+ c
Question 34. ∫x3 sin⁡x4 dx
Solution:
Given that I = ∫x3sin⁡x4 dx   …..(i)
Let us considered x= t then,
On differentiating both side we get,
d(x4) = dt
4x3dx = dt
x= dt/4
Now on putting x= 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⁡x+ c
Question 35. ∫(xsin-1x2)/√(1 – x4) dx
Solution:
Given that I = ∫(xsin-1⁡x2)/√(1 – x4) dx  …….(i)
Let us considered sin-1x= 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-1x= 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)+ c
Hence, I = 1/4 (sin-1x2)+ c
Question 36. ∫x3sin⁡(x+ 1)dx
Solution:
Given that I = ∫x3 sin⁡(x+ 1)dx ……..(i)
Let us considered x+ 1 = t then,
On differentiating both side we get,
d(x+ 1) = dt
x3 dx = dt/4
Now on putting x+ 1 = t and x3dx = dt/4 in equation (i), we get
I = ∫ sin⁡t dt/4
= -1/4 cos⁡t + c
= -1/4 cos⁡(x+ 1) + c
Hence, I = -1/4 cos⁡(x+ 1) + c
Question 37. ∫(x + 1)ex/(cos2⁡(xex) dx
Solution:
Given that I = ∫((x + 1)ex)/(cos2⁡(xex)) dx   ……(i)
Let us considered xe= t then, 
On differentiating both side we get,
d(xex) = dt
(e+ xex)dx = dt
(x + 1)exdx = dt
Now on putting xe= t and (x + 1)exdx = dt in equation (i), we get
I = ∫dt/(cos2⁡t)
= ∫ sec2tdt
= tan⁡t + c
= tan⁡(xex) + c
Hence, I = tan⁡(xex) + c
Question 38. ∫x^2 e^{x^3} cos⁡(e^{x^3})dx
Solution:
Given that I = ∫x^2 e^{x^3} cos⁡(e^{x^3})dx   ……..(i)
Let us considered e^{x^3} = t then,
On differentiating both side we get,
d(e^{x^3}) = dt
3x2e^{x^3}dx = dt
x2 e^{x^3}dx = dt/3
Now on putting e^{x^3} = t and x2e^{x^3} dx = dt/3 in equation (i), we get
I = ∫cos⁡t dt/3
= (sin⁡t)/3 + c
Hence, I = sin⁡(e^{x^3})/3 + c
Question 39. ∫2xsec3(x+ 3)tan⁡(x+ 3)dx
Solution:
Given that I = ∫2xsec3(x+ 3)tan⁡(x+ 3)dx    ………(i)
Let us considered sec⁡(x+ 3) = t then,
On differentiating both side we get,
d[sec⁡(x+ 3)] = dt
2xsec⁡(x+ 3)tan⁡(x+ 3)dx = dt
Now on putting sec⁡(x+ 3) = t and 2xsec⁡(x+ 3)tan⁡(x+ 3)dx = dt in equation (i), we get
I = ∫t2 dt
= t3/3 + c
= 1/3 [sec⁡(x+ 3)]+ c
Hence, I = 1/3 [sec⁡(x+ 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)+ 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 ∫t1/2 dt
=-1/2×t3/2/(3/2) + c
=-1/3 t3/2 + 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)= 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/xdx = -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 + t– 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)ex)/(sin2(xex)) dx
Solution:
Given that I = ∫((x + 1)ex)/(sin2⁡(xex)) dx   …….(i)
Let us considered xe= t then,
d(xex) = dt
(xe+ ex)dx = dt
(x + 1)exdx = dt
Now on putting xe= t and (x + 1)ex dx = dt in equation (i), we get
I = ∫dt/(sin2t)
= ∫cos⁡ec2t dt
= -cot + c
Hence, I = -cot⁡(xex) + c

Question 49. ∫5^{x + tan^{-1}x}(\frac{x^2 + 2}{x^2 + 1})dx

Solution:

Given that I = ∫5^{x + tan^{-1}x}(\frac{x^2 + 2}{x^2 + 1})dx  ……..(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 + x+ 1)/(1 + x2))dx = dt

((x+ 2))/((x+ 1)) dx = dt

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

I = ∫5tdt

= 5t/(log⁡5) + c

5^{x + tan^{-1}x}/(log⁡5) + c

Hence, I = 5^{x + tan^{-1}x}/(log⁡5) + c

Question 50. ∫\frac{e^{msin^{-1}x}}{\sqrt{1-x^2}} dx

Solution:

Given that I = ∫\frac{e^{msin^{-1}x}}{\sqrt{1-x^2}} dx  ……(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 = ∫etdt/m

= 1/m et+c

\frac{1}{m}e^{msin^{-1}x}+c

Hence, I = \frac{1}{m}e^{msin^{-1}x}+c

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. ∫\frac{e^{mtan^{-1}x}}{(1+x^2)}dx

Solution:

Given that I = ∫\frac{e^{mtan^{-1}x}}{(1+x^2)}dx

Let us considered tan-1⁡x = t, then

On differentiating the above function we have, 

1/(1 + x2) dx = dt

∫\frac{e^{mtan^{-1}x}}{(1+x^2 )} dx  = ∫emt × dt

=∫\frac{e^{mtan^{-1}x}}{(1+x^2 )} dx = emt/m

On Substituting the value of t, we 

I = \frac{e^{mtan^{-1}x}}{(m)} + c

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

Solution:

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

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

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

= ∫ x/(2a2) (√(x+ a2) – √(x– a2))dx

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

Let us considered x= t then, 

On differentiating the above function we have, 

d(x2) = dt

2xdx = dt

xdx = dt/2

Now on putting x= 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-1x= 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⁡x= 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)+ 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)+ 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. ∫xe^{x^{2}} dx

Solution:

Given that I = ∫xe^{x^{2}} 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 x= t and xdx = dt/2 in equation (i), we get

I = ∫etdt/2

= 1/2 et+c

= 1/2 e^{x^{2}} + c

Hence, I = 1/2 e^{x^{2}} + c

Question 60. ∫e2x/(1 + ex) dx

Solution:

Given that I = ∫e2x/(1 + ex) dx  …….(i)

Let us considered 1 + e= t then, 

On differentiating the above function we have, 

d(1 + ex) = dt

exdx = dt

dx = dt/ex

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

I = ∫e2x/t × dt/ex 

= ∫ex/t dt

= ∫ (t – 1)/t dt

= ∫ (t/t – 1/t)dt

= t – log⁡|t| + c

= (1 + ex) – log⁡|1 + ex| + c

Hence, I = 1 + e– log⁡|1 + ex| + 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∫((t– 1) + t)/((t– 1) + 2) × tdt   [x + 1 = t2]

= 2∫(t+ t – 1)/(t+ 1) tdt

= 2∫ (t+ t– t)/(t2 + 1) dt

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

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

Let I= ∫t3/(t+ 1) dt

I= ∫t2/(t+ 1) dt

and I= ∫t/(t+ 1) dt

Now, I= ∫t3/(t+ 1) dt

= ∫(t – t/(t+ 1))dt

= t2/2 – 1/2 log⁡(t+ 1)      

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

Since, I= ∫t2/(t+ 1) dt

 = ∫ (t+ 1 – 1)/(t+ 1) dt

 = ∫(t+ 1)/(t+ 1) dt – ∫1/(t+ 1) dt

 = ∫dt – ∫1/(t+ 1) dt

I= t – tan-1⁡(t2) + c2 ………….(iv)

and, 

I= ∫t/(t+ 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⁡(t+ 1) + c+ t-tan-1⁡(t2) + c– 1/2 log⁡(1 + t2) + c3]

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

= 2[t2/2 + t – tan-1⁡(t2) – log⁡(1 + t2) + c4  [Putting c+ c+ c= c4]

= t+ 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 2c= c]

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

Question 64. ∫5^{5^{5^{x}}} 5^{5^x} 5^x dx

Solution:

Given that I = ∫5^{5^{5^{x}}} 5^{5^x} 5^x dx  …….(1)

Let us considered 5^{5^{5^{x}}} = t then 

On differentiating the above function we have, 

 d(5^{5^{5^{x}}}) = dt

5^{5^{5^{x}}} × 5^{5^x} × 5x × (log⁡5)3 dx = dt

5^{5^{5^{x}}} 5^{5^x} 5^x dx = dt/((log⁡5)^3 ))

Now on putting 5^{5^{5^{x}}} = t and 5^{5^{5^{x}}} 5^{5^x} 5^x 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 = 5^{5^{5^{x}}}/((log⁡5)3) + c)

Question 65. ∫1/(x√(x– 1)) dx

Solution:

Given that I = ∫1/(x√(x– 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√(t– 1)) × dt/2x

= 1/2 ∫ 1/(x2 √(t– 1)) dt

= 1/2 ∫ 1/(t√(t– 1)) dt

= 1/2 sec-1t + c

= 1/2 sec-1x+ c

Hence, I = 1/2 sec-1x2 + c

Question 66. ∫√(e– 1) dx

Solution:

Given that I = ∫√(e– 1) dx   ……..(i)

Let us considered e– 1 = t2 then, 

On differentiating the above function we have, 

d(e– 1) = dt(t2)

ex dx = 2tdt

dx = 2t/edt

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

Now on putting e– 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)/(t+ 1) – 1/(t+ 1)]dt

= 2∫dt – 2∫1/(t2 + 1) dt

= 2t – 2tan-1⁡(t) + c

= 2√(e– 1) – 2tan-1(√(e– 1)) + c

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

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

Solution:

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

= ∫1/(x + 1)((x + 1)+ 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 = √(x+ 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)/√(x+ 2x + 2)]

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

Question 68. ∫x5/√(1 + x3) dx

Solution:

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

Let us considered 1 + x= 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 + x= t2 and dx = 2tdt/3x2 in equation (i), we get

I = ∫ x5/√t2 * 2t/3xdt

= ∫x5/t * 2t/3xdt

= 2/3∫x3 dt

= 2/3 ∫( t– 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 – x= tthen, 

On differentiating the above function we have, 

d(5 – x2) = t2

-2xdx = 2tdt

dx = (-t)/x dt

Now on putting 5 – x= 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 – x= t2]

= -4∫(5t– t5)dt

= -20×t3/3 + 4 t5/5 + c

= (-20)/3 × t+ 4/5 × t+ 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 (x+ 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 (x+ 1)-3/4/(x2x-3))

= (x+ 1)-3/4/(x5(x4)-3/4

= 1/x5 ((x+ 1)/x4)-3/4

= 1/x5 (1 + 1/x4)-3/4

Let us considered 1/x4 = t

-4/xdx = dt

1/xdx = -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)/t× -dt/(sin⁡x)

= -∫(sin4⁡x)/t4 dt

= -∫(1 – cos2x)2/tdt

= -∫(1 – t2)2/tdt

= -∫(1 + t– 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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