RD Sharma Class 12 Ex 19.25 Solutions Chapter 19 Indefinite Integrals

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.25
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 19.25 Solutions Chapter 19 Indefinite Integrals

Evaluate the following integrals:

Question 1. ∫x cos⁡xdx

Solution:

Given that, I = ∫x cos⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x∫cos⁡xdx – ∫(1 × ∫cos⁡xdx)dx + c

= xsin⁡x – ∫sin⁡xdx + c

Hence, I = x sin⁡x + cos⁡x + c

Question 2. ∫log⁡(x + 1)dx

Solution:

Given that, I = ∫log⁡(x + 1)dx

= ∫1 × log⁡(x + 1)dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = log⁡(x + 1)∫1dx – ∫(1/(x + 1) × ∫ 1dx)dx + c

= xlog⁡(x + 1) – ∫(x/(x + 1))dx + c

= x log⁡(x + 1) – ∫(1 – 1/(x + 1))dx + c

Hence, I = x log⁡(x + 1) – x + log⁡(x + 1) + c

Question 3. ∫x3 log⁡xdx

Solution:

Given that, I = ∫ x3 log⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = log⁡x ∫x3 dx – ∫(1/x × ∫x3 dx)dx + c

= x4/4 log⁡x – ∫x4/4x dx+c

= x4/4 log⁡x – 1/4∫x3 dx + c

= x4/4 log⁡x – 1/4 ∫x4/4 dx + c

I = x4/4 log⁡x – 1/16 x+ c

Question 4. ∫xex dx

Solution:

Given that I = ∫xex dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = xe– ∫1.ex dx

= xe– e+ c

Hence, I = = xe– e+ c

Question 5. ∫xe2x dx

Solution:

Given that, I = ∫xe2x dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x∫e2x dx – ∫(1 × ∫ e2x dx) dx + c

= x∫e2x dx – ∫(1 × ∫e2x dx)dx + c

= (xe2x)/2 – ∫(e2x/2)dx + c

= (xe2x)/2 – e2x/4 + c

Hence, I = (x/2 – 1/4) e2x + c

Question 6. ∫x2 e-x dx

Solution:

Given that I = ∫x2 e-x dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x2 ∫e-x dx – ∫(2x∫e-x dx)dx

= -x2 e-x – ∫(2x)(-e-x)dx

= -x2 e-x + 2∫xe-x dx

= -x2 e-x + 2[x∫e-x dx – ∫(1 × ∫ e-x dx) dx]

= -x2 e-x + 2[x(-e-x) – ∫(-e-x)dx]

= -x2 e-x – 2xe-x + 2∫e-x dx

Hence, I = -x2 e-x – 2xe-x – 2e-x + c

Question 7.  ∫ x2cos⁡xdx

Solution:

Given that, I = ∫ x2cos⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

 I = x2 ∫ cos⁡xdx – ∫(2x)cos⁡xdx)dx

= x2 sin⁡x – 2∫(x)(sin⁡x)dx

= x2 sin⁡x – 2[x∫sin⁡xdx – ∫(1 × ∫sin⁡xdx)dx]

= x2 sin⁡x – 2[x(-cos⁡x) – ∫(-cos⁡x)dx]

= x2 sin⁡x + 2xcos⁡x – 2∫(cos⁡x)dx

Hence, I = x2sin⁡x + 2xcos⁡x – 2sin⁡x + c

Question 8. ∫x2cos⁡2xdx

Solution:

Given that, I = ∫x2cos⁡2xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x2 ∫cos⁡2xdx – ∫(2x∫ cos⁡2xdx)dx

= x2 (sin⁡2x)/2 – 2∫x((sin⁡2x)/2)dx

= 1/2 x2 sin⁡2x – ∫xsin⁡2xdx

= 1/2 x2 sin⁡2x – [x∫sin⁡2xdx – ∫ (1∫ sin⁡2xdx)dx]

= 1/2 x2 sin⁡2x – [x((-cos⁡2x)/2) – ∫(-(cos⁡2x)/2)dx]

= 1/2 x2sin⁡2x + x/2 cos⁡2x – 1/2 ∫(cos⁡2x)dx

Hence, I = 1/2 x2 sin⁡2x + x/2 cos⁡2x – 1/4 sin⁡2x + c

Question 9. ∫xsin⁡2xdx

Solution:

Given that, I =∫xsin⁡2xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

 I = x∫sin⁡2xdx – ∫(1)sin⁡2xdx)dx

= x(-(cos⁡2x)/2) – ∫(-(cos⁡2x)/2)dx

= -x/2 cos⁡2x + 1/2 ∫cos⁡2xdx

= -x/2 cos⁡2x + 1/2(sin⁡2x)/2 + c

Hence, I = -x/2 cos⁡2x + 1/4 sin⁡2x + c

Question 10. ∫(log⁡(log⁡x))/x dx

Solution:

 Given that, I = ∫(log⁡(log⁡x))/x dx 

= ∫(1/x)(log⁡(log⁡x))dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = log⁡log⁡x]1/x dx – ∫(1/(xlog⁡x)∫1/x dx)dx

= log⁡x × log⁡(log⁡x) – ∫(1/(xlog⁡x) log⁡x)dx

= log⁡x × log⁡(log⁡x) – ∫1/x dx

= log⁡x × log⁡(log⁡x) – log⁡x + c

Hence, I = log⁡x(log⁡log⁡x – 1) + c

Question 11. ∫x2 cos⁡xdx

Solution:

Given that I = ∫xcos⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x2∫ cos⁡xdx – ∫(2x]cos⁡xdx)dx

= x2sin⁡x – 2∫xsin⁡xdx

= x2 sin⁡x – 2[x∫sin⁡xdx – ∫(1]sin⁡xdx)dx]

= x2 sin⁡x – 2[x(-cos⁡x) – ∫(-cos⁡x)dx]

= x2 sin⁡x + 2xcos⁡x – 2∫(cos⁡x)dx

Hence, I = x2 sin⁡x + 2xcos⁡x – 2sin⁡x + c

Question 12. ∫xcosec2⁡xdx

Solution :

Given that, I = ∫xcosec2⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x∫cosec2xdx – ∫(∫ cosec2xdx)dx

= -xcot⁡x + ∫cot⁡xdx

= -x cot⁡x + log ⁡|sin⁡x| + c

Hence, I = -x cot⁡x + log ⁡|sin⁡x| + c

Question 13. ∫xcos2xdx

Solution:

Given that, I = ∫xcos2xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x∫cos2⁡xdx – ∫(1∫ cos2xdx)dx

= x∫((cos⁡2x + 1)/2)dx – ∫(∫((1 + cos⁡2x)/2)dx)dx

= x/2 [(sin⁡2x)/2 + x] – 1/2∫(x + (sin⁡2x)/2)dx

= x/4 sin⁡2x + x2/2 – 1/2 × x2/2 – 1/4 (-(cos⁡2x)/2) + c

Hence, I = x/4 sin⁡2x + x2/4 + 1/8 cos⁡2x + c

Question 14. ∫xn log⁡x dx

Solution:

Given that, I = ∫xn log⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = log⁡x∫xn dx – ∫(1/x ∫xndx)dx

= xn+1/(n + 1) log⁡x – ∫(1/x × xn+1/(n + 1))dx

= xn+1/(n + 1) log⁡x – ∫(xn/(n + 1))dx

Hence, I = xn+1/(n + 1) log⁡x – 1/(n + 1)2 × (xn+1) + c

Question 15. ∫(log⁡x)/xn dx

Solution:

Given that, I = ∫(log⁡x)/xdx = ∫(log⁡x)(1/xn)dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = log⁡x∫(1/xn)dx – ∫((d(log⁡x))/dx)(∫(1/xn)dx)dx

= log⁡x(x1-n/(1 – n)) – ∫1/x (x1-n/(1 – n))dx

= log⁡x(x1-n/(1 – n)) – ∫(xn/(1 – n))dx

= log⁡x(x1-n/(1 – n)) – (1/(1 – n))(x1-n/(1 – n))

Hence, I = log⁡x(x1-n/(1 – n)) – (x1-n/([1 – n]2)) + c

Question 16. ∫x2 sin2⁡xdx

Solution:

Given that, I = ∫x2 sin2⁡xdx

= ∫x2 ((1 – cos⁡2x)/2)dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= ∫x2/2 dx – ∫((x2 cos⁡2x)/2)dx

= x3/6 – 1/2 [∫x2 cos⁡2xdx]

= x3/6 – 1/2 [x2 ∫cos⁡2xdx – ∫ (2x∫cos⁡2xdx)dx]

= x3/6 – 1/2 (x2(sin⁡2x)/2) + 1/2 × 2∫(x (sin⁡2x)/2)dx

= x3/6 – 1/4 x2sin⁡2x + 1/2 [x ∫sin⁡2xdx – ∫(1∫sin⁡2xdx)dx]

= x3/6 – 1/4 x2 sin⁡2x + 1/2 [x(-(cos⁡2x)/2) – ∫(-(cos⁡2x)/2)dx] 

= x3/6 – 1/4 x2 sin⁡2x + 1/2 x(-(cos⁡2x)/2) + 1/4 × (sin2x/2) + c

= x3/6 – 1/4 x2 sin⁡2x – 1/4 x(cos⁡2x) + 1/8 × (sin2x) + c

Hence, I = x3/6 – 1/4 x2 sin⁡2x – 1/4 x(cos⁡2x) + 1/8 × (sin2x) + c

Question 17. ∫2x^3 e^{x^2} xdx

Solution:

Given that, l = ∫2x^3 e^{x^2} xdx

 Let us assume, x2 = t

2xdx = dt

I = ∫t × et dt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= t∫et dt – ∫(1 × ∫etdt)dt

= te– ∫et dt

= te– e+ c

= et-1 + c

Hence, I = e^{x^2} (x2 – 1) + c

Question 18. ∫x3 cos⁡x2 dx

Solution:

Given that, I = ∫x3 cos⁡x2 dx

Let us assume x= t

2xdx = dt

I = 1/2 ∫tcos⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 1/2[t∫cos⁡tdt – ∫(1 × ∫cos⁡tdt)dt]

= 1/2 [t × sin⁡t – ∫sin⁡tdt]

= 1/2[tsin⁡t + cos⁡t] + c

Hence, I = 1/2 [x² sin⁡x+ cos⁡x2] + c

Question 19. ∫xsin⁡xcos⁡xdx

Solution:

Given that, I = ∫xsin⁡xcos⁡xdx

 = ∫x/2(2sin⁡xcos⁡x)dx

 = 1/2 ∫xsin⁡2xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 1/2 [x∫sin⁡2xdx – ∫(1 × ∫sin⁡2xdx)dx]

= 1/2 [x((-cos⁡2x)/2) – ∫((-cos⁡2x)/2)dx]

= -1/4 xcos⁡2x + 1/4 ∫cos⁡2xdx

Hence, I = -1/4 xcos⁡2x + 1/8 sin⁡2x + c

Question 20. ∫sin⁡x(log⁡cos⁡x)dx

Solution:

Given that, I = ∫sin⁡x(log⁡cos⁡x)dx

 Let us considered, cos⁡x = t

 -sin⁡xdx = dt

I = -∫ log⁡tdt

 = -∫1 × log⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= -[log⁡t∫dt – ∫(1/t × ∫dt)dt]

= -[tlog⁡t – ∫1/t × tdt]

= -[tlog⁡t-∫  dt]

= -[tlog⁡t – t + c1 ]

= t(1 – logt) + c

Hence, I = cosx(1 – logcosx) + c 

Evaluate the following integrals:

Question 21. ∫(log⁡x)2 x dx

Solution:

Given that, I = ∫(log⁡x)2 x dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = (log⁡x)2∫xdx – ∫(2(log⁡x)(1/x) ∫xdx)dx

= x2/2(log⁡x)– 2∫(log⁡x)(1/x)(x2/2)dx

= x2/2(log⁡x)– ∫x(log⁡x)dx

= x2/2(log⁡x)– [log⁡x∫xdx – ∫ (1/x ∫xdx)dx]

= x2/2(log⁡x)– [x22/2 log⁡x – ∫(1/x × x2/2)dx]

= x2/2(log⁡x)– x2/2 log⁡x + 1/2 ∫xdx

= x2/2(log⁡x)– x2/2 log⁡x + 1/4 x+ c

Hence, I = x2/2 [(log⁡x)– log⁡x + 1/2] + c

Question 22. ∫e√x dx

Solution:

Given that, I = ∫ e√x dx

 Let us assume, √x = t

x = t2

dx = 2tdt

I = 2∫ et tdt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = 2[t∫etdt – ∫(1∫etdt)dt]

= 2[te– ∫et dt]

= 2[te– et] + c

= 2et (t – 1) + c

Hence, I = 2e√x(√x – 1) + c

Question 23. ∫(log⁡(x + 2))/((x + 2)2) dx

Solution:

Given that, I = ∫(log⁡(x + 2))/((x + 2)2) dx

 Let us assume (1/(x + 2) = t

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

I = -∫log⁡(1/t)dt

= -∫log⁡t-1 dt

= -∫1 × log⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = log⁡t∫dt – ∫(1/t ∫dt)dt

= tlog⁡t – ∫(1/t × t)dt

= tlog⁡t – ∫dt

= tlog⁡t – t + c

= 1/(x + 2) (log⁡(x + 2)-1 – 1) + c

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

Question 24. ∫(x + sin⁡x)/(1 + cos⁡x) dx

Solution:

Given that, I = ∫(x + sin⁡x)/(1 + cos⁡x) dx

= ∫x/(2cos2x/2) dx + ∫(2sin⁡x/2 cos⁡x/2)/(2cos2⁡x/2) dx

= 1/2 ∫xsec2⁡x/2 dx + ∫tan⁡x/2 dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 1/2 [x∫sec2x/2 dx – ∫(1∫ sec²x/2 dx)dx] + ∫tan⁡x/2 dx

= 1/2 [2xtan⁡x/2 – 2∫tan⁡x/2 dx] + ∫tan⁡x/2 dx + c

= xtan⁡x/2 – ∫tan⁡x/2 dx + ∫tan⁡x/2 dx+c

Hence, I = xtan⁡x/2 + c

Question 25. ∫log10xdx

Solution:

Given that, I = ∫log10⁡xdx

= ∫(log⁡x)/(log⁡10) dx

= 1/(log⁡10) ∫1 × log⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 1/(log⁡10) [log⁡x∫dx – ∫(1/x ∫dx)dx]

= 1/(log⁡10) [xlog⁡x – ∫(x/x)dx]

= 1/(log⁡10)[xlog⁡x – x]

Hence, I = (x/(log⁡10)) × (log⁡x – 1) + c

Question 26. ∫cos⁡√x dx

Solution:

Given that, I = ∫cos⁡√x dx

Let us assume, √x = t

x = t2

dx = 2tdt

= ∫2tcos⁡tdt

I = 2∫tcos⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = 2[t]cos⁡tdt – ∫(1 ∫cos⁡tdt)dt]

= 2[tsin⁡t – ∫sin⁡tdt]

= 2[tsin⁡t + cos⁡t] + c

Hence, I = 2[√x sin⁡√x + cos√x] + c

Question 27. ∫(xcos-1x)/√(1 – x2) dx

Solution:

Given that, I = ∫(xcos-1x)/√(1 – x2) dx

Let us assume, t = cos-1⁡x

dt = (-1)/√(1 – x2) dx

Also, cost = x

I = -∫tcos⁡tdt

Now, using integration by parts,       

So, let

u = t;

du = dt

∫cos⁡tdt = ∫dv

sin⁡t = v

Therefore,

 I = -[tsint – ∫sin⁡tdt]

= -[tsint + cos⁡t] + c

On substituting the value t = cos-1x we get,

 I = -[cos-1⁡xsin⁡t + x] + c

Hence, I = -[cos-1⁡x√(1 – x²) + x] + c

Question 28. ∫cosec3xdx

Solution:

Given that, I =∫cosec3xdx

 =∫cosec⁡x × cosec2xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= cosec⁡x × ∫cosec2xdx + ∫(cosec⁡xcot⁡x]cosec2xdx)dx

= cosecx × (-cot⁡x) + ∫cosec⁡xcot⁡x(-cot⁡x)dx

= -cosec⁡xcot⁡x – ∫cosecx⁡cot2⁡xdx

= -cosec⁡xcot⁡x – ∫cosec⁡x(cosec2x – 1)dx

= -cosec⁡xcot⁡x – ∫cosec3xdx + ∫cosecxd⁡x

I = -cosec⁡xcot⁡x – I + log⁡|tan⁡x/2| + c1

2l = -cosec⁡xcot⁡x + log⁡|tan⁡x/2| + c1

Hence, I = -1/2cosecx⁡cot⁡x + 1/2 log⁡|tan⁡x/2| + c

Question 29. ∫sec-1√x dx

Solution:

Given that, I = ∫sec-1⁡√x dx

 Let us assume, √x = t

x = t2

dx = 2tdt

I = ∫2tsec-1⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 2[sec-1⁡t∫tdt – ∫(1/(t√(t2-1))∫tdt)dt]

= 2[t2/2 sec-1 – ∫(t/(2t√(t– 1)))dt]

= t2 sec-1⁡t – ∫t/√(t– 1) dt

= t2 sec-1⁡t – 1/2∫2t/√(t– 1) dt

= t2 sec-1⁡t – 1/2 × 2√(t– 1) + c

Hence, I = xsec-1⁡√x – √(x – 1) + c

Question 30. ∫sin-1√x dx

Solution:

Given that, I = ∫sin-1⁡√x dx 

Let us assume, x = t

dx = 2tdt

∫sin-1√x dx = ∫sin-1⁡√(t2) 2tdt

= ∫sin-1⁡t2tdt

= sin⁡-1t∫2tdt – (∫(dsin-1⁡t)/dt (∫2tdt)dt

= sin-1⁡t(t2) – ∫1/√(1 – t2) (t2)dt

Now, lets solve ∫1/√(1 – t2) (t2)dt

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

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

As we know that, value of ∫1/√(1 – t2) dt = sin-1⁡t

So, the remaining integral to evaluate is 

∫(t– 1)/√(1 – t2) dt= ∫-√(1 – t2) dt

Now, substitute, t = sin⁡u, dt = cos⁡udu, we gte

∫-√(1 – t2) dt = ∫-cos2udu = -∫[(1 + cos⁡2u)/2]du

= -u/2-(sin⁡2u)/4

Now substitute back u = sin-1x and t = √x, we get

= -(sin-1√x)/2 – (sin⁡(2sin-1⁡√x))/4

∫sin-1√x dx = xsin-1⁡√x-(sin-1⁡√x)/2 – (sin⁡(2sin-1√x))/4

sin⁡(2sin-1⁡√x) = 2√x √(1 – x)

Hence, I = xsin-1⁡√x – (sin-1⁡√x)/2 – √(x(1 – x))/2 + c

Question 31. ∫xtan2⁡xdx

Solution:

Given that, I =∫xtan2⁡xdx

 = ∫x(sec2x – 1)dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

 = ∫xsec8xdx – ∫xdx

 = [x∫sec2xdx – ∫(1∫sec2xdx)dx] – x2/2

 = xtan⁡x – ∫tan⁡xdx – x2/2

Hence, I = xtan⁡x – log⁡|sec⁡x| – x2/2 + c

Question 32. ∫ x((sec⁡2x – 1)/(sec⁡2x + 1))dx

Solution:

Given that, I = ∫ x((sec⁡2x – 1)/(sec⁡2x + 1))dx

= ∫x((1 – cos⁡2x)/(1 + cos⁡2x))dx

= ∫x((sec2⁡x)/(cos2⁡x))dx

= ∫xtan2xdx

= ∫x(sec2⁡x – 1)dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= ∫xsec2xdx – ∫dx

= [x∫sec2⁡xdx – ∫(1∫  sec2xdx)dx] – x2/2

= xtan⁡x – ∫tan⁡xdx – x2/2

= xtan⁡x – log|secx| – x2/2 + c

Hence, I = xtan⁡x – log|secx| – x2/2 + c

Question 33. ∫(x + 1)exlog⁡(xex)dx

Solution:

Given that, I = ∫(x + 1)exlog⁡(xex)dx

Let us assume, xe= t

(1 × e+ xex)dx = dt

(x + 1)exdx = dt

I = ∫log⁡tdt

= ∫1 × log⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= log⁡t∫dt – ∫(1/t∫dt)dt

= tlog⁡t – ∫(1/t × t)dt

= tlog⁡t – ∫dt

= tlog⁡t – t + c

= t(log⁡t – 1) + c

Hence, I = xex (log⁡xe– 1) + c

Question 34. ∫sin-1(3x – 4x3)dx

Solution:

Given that, I = ∫sin-1(3x – 4x3)dx

Let us assume, x = sin⁡θ

dx = cos⁡θdθ

= ∫sin-1⁡(3sin⁡θ – 4sin3⁡θ)cos⁡θdθ

= ∫sin-1(sin⁡3θ)cos⁡θdθ

= ∫3θcos⁡θdθ

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 3[θ]cos⁡θdθ – ∫(1∫cos⁡θdθ)dθ]

= 3[θsin⁡θ – ∫sin⁡θdθ]

= 3[θsin⁡θ + cos⁡θ] + c

Hence, I = 3[xsin-1⁡x + √(1 – x2)] + c)

Question 35. ∫sin-1(2x/(1 + x2))dx.

Solution:

Given that, I = ∫sin-1(2x/(1 + x2))dx

Let us assume, x = tan⁡θ

dx = sec2⁡θdθ

sin-1⁡(2x/(1 + x2)) = sin-1⁡((2tan⁡θ)/(1 + tan²⁡θ))

= sin-1⁡(sin⁡2θ) = 2θ

∫sin-1⁡(2x/(1 + x2))dx = ∫2θsec2θdθ = 2∫θsec2⁡θdθ

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

2[θ∫sec2⁡θdθ – ∫{(d/dθ θ) ∫sec2θdθ}dθ

= 2[θtan⁡θ – ∫tan⁡θdθ]

= 2[θtan⁡θ + log⁡|cos⁡θ|] + c

= 2[xtan-1⁡x + log⁡|1/√(1 + x2)|] + c

= 2xtan-1⁡x + 2log⁡(1 + x2)1/2 + c

= 2xtan-1⁡x + 2[-1/2 log⁡(1 + x2)] + c

= 2xtan-1⁡x – log⁡(1 + x2) + c

Hence, I = 2xtan-1⁡x – log⁡(1 + x2) + c

Question 36. ∫ tan-1((3x – x3)/(1 – 3x2))dx

Solution:

Given that, I = ∫tan-1⁡((3x – x3)/(1 – 3x2))dx

 Let us assume, x = tan⁡θ

dx = sec2⁡θdθ

I = ∫tan-1⁡((3tan⁡θ – tan3θ)/(1 – 3tan2⁡x)) sec2⁡θdθ

=∫tan-1⁡(tan⁡3θ)sec2θdθ

= ∫3θsec2θdθ

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 3[θ∫ sec2⁡θdθ – ∫(1∫sec2θdθ)dθ]

= 3[θtan⁡θ – ∫tan⁡θdθ]

= 3[θtan⁡θ + log⁡sec⁡θ] + c

= 3[xtan-1x – log⁡√(1 + x2)] + c

Hence, I = 3[xtan-1x – log⁡√(1 + x2)] + c

Question 37. ∫x2sin-1xdx

Solution:

Given that, I = ∫x2sin-1⁡xdx

I = sin-1x∫x2 dx – ∫(1/√(1 – x2) ∫x2 dx)dx

= x3/3 sin-1⁡x – ∫x3/(3√(1 – x2)) dx

I = x3/3 sin-1⁡x – 1/3 I+ c…..(1)

Let I= ∫x3/√(1 – x2) dx

Let 1 – x= t2

-2xdx = 2tdt

-xdx = tdt

I= -∫(1 – t2)tdt/t

= ∫(t– 1)dt

= t3/3 – t + c2

= (1 – x2)3/2/3 – (1 – x2)1/2 + c2

Now, put the value of I1 in eq(1), we get

Hence, I = x3/3 sin-1⁡x – 1/9 (1 – x2)3/2 + 1/3 (1 – x2)1/2 + c

Question 38. ∫(sin-1x)/x2dx

Solution:

Given that, I =∫(sin-1⁡x)/x2dx

 = ∫(1/x2)(sin-1⁡x)dx

I = [sin-1⁡x∫1/x2dx – ∫(1/√(1 – x2) ∫1/x2dx)dx]

= sin-1×(-1/x) – ∫1/√(1 – x2) (-1/x)dx

I = -1/x sin-1x + ∫1/(x√(1 – x2)) dx

I = -1/x sin-1⁡x + I1  …….(1)

Where,

I= ∫1/(x√(1 – x2)) dx

Let 1 – x= t2

-2xdx = 2tdt

I= ∫x/(x2√(1 – x2)) dx

= -∫tdt/((1 – t2) √t)

= -∫dt/((1 – t2))

= ∫1/(t– 1) dt

= 1/2 log⁡|(t – 1)/(t + 1)| 

= 1/2 log⁡|(t – 1)/(t + 1)|

= 1/2 log⁡|(√(1 – x2) – 1)/(√(1 – x2) + 1)| + c1

Now, put the value of I1 in eq(1), we get

I = -(sin-1x)/x + 1/2 log⁡|((√(1 – x2) – 1)/(√(1 – x2) + 1))((√(1 – x2) – 1)/(√(1 – x2) – 1))| + c 

= -(sin-1⁡x)/x + 1/2 log⁡|(√(1 – x2) – 1)2/(1 – x– 1)| + c

= -(sin-1⁡x)/x + 1/2 log⁡|(√(1 – x2) – 1)2/(-x2)| + c

= -(sin-1⁡x)/x + log⁡|(√(1 – x2) – 1)/(-x)| + c

Hence, I = -(sin-1x)/x + log⁡|(1 – √(1 – x2))/x| + c

Question 39. ∫(x2 tan-1x)/(1 + x2) dx

Solution:

Given that, I = ∫(x2 tan-1⁡x)/(1 + x2) dx

Let us assume, tan-1⁡x = t     [x = tan⁡t]

1/(1 + x2) dx = dt

I = ∫t × tan2tdt

= ∫t(sec2⁡t – 1)dt

= ∫(tsec2t – t)dt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= ∫tsec2tdt – ∫tdt

= [t∫sec2tdt – ∫(1)sec2⁡tdt)dt] – t2/2

= [t × tan⁡t – ∫tan⁡tdt] – t2/2

= t tan⁡t – log⁡sec⁡t – t2/2 + c

= xtan-1⁡x – log⁡√(1 + x2) – (tan2x)/2 + c

Hence, I = xtan-1⁡x – 1/2 log⁡|1 + x2| – (tan2⁡x)/2 + c

Question 40. ∫cos-1(4x– 3x)dx

Solution:

Given that, I = ∫cos-1⁡(4x– 3x)dx

 Let us assume, x = cos⁡θ

dx = -sin⁡θdθ

I = -∫cos-1(4cos⁡3θ – 3cos⁡θ)sin⁡θdθ

= – ∫cos-1(cos⁡3θ)sin⁡θdθ

= -∫3θsin⁡θdθ

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= -3[θ]sin⁡θdθ – ∫(1∫sin⁡θdθ)dθ]

= -3[-θcos⁡θ + ∫cos⁡θdθ]

= 3θcos⁡θ – 3sin⁡θ + c

Hence, I = 3xcos-1x – 3√(1 – x2) + c

Evaluate the following integrals:

Question 41. ∫cos-1⁡((1 – x2)/(1 + x2))dx 

Solution:

Given that, I = ∫cos-1⁡((1 – x2)/(1 + x2))dx)

Let us considered x = tan⁡t

dx = sec²tdt

I = ∫cos-1⁡((1 – tan2t)/(1 + tan2⁡t)) sec2tdt

= ∫cos-1(cos⁡2t)sec2tdt

= ∫2tsec2⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = 2[t∫sce2tdt – ∫(1∫sec2⁡tdt)dt]

= 2[t × tan2t – ∫tan⁡tdt]

= 2[t × tan2t – log⁡sec⁡t] + c

= 2[xtan-1x – log⁡√(1 + x2)] + c

Hence, I = 2xtan-1x – log⁡|1 + x2| + c

Question 42. ∫tan-1⁡(2x/(1 – x2))dx

Solution:

Given that, I = ∫tan-1⁡(2x/(1 – x2))dx

Let us considered x = tan⁡θ

dx = sec2θdθ

I = ∫tan-1⁡((2tan⁡θ)/(1 – tan2θ)) sec2θdθ

= ∫tan-1⁡(tan⁡2θ)sec2θdθ

= ∫2θsec2θdθ

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = 2[θ∫sec2θdθ – ∫(1∫ sec2⁡θdθ)dθ]

= 2[θtan⁡θ – ∫tan⁡θdθ]

= 2[θtan⁡θ – log⁡sec⁡θ] + c

= 2[xtan-1⁡x – log⁡√(1 + x2)] + c

Hence, I = 2xtan-1⁡x – log⁡|1 + x2| + c

Question 43. ∫(x + 1)log⁡xdx

Solution:

Given that, I = ∫(x + 1)log⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = log⁡x∫ (x + 1)dx – ∫(1/x ∫(x + 1)dx)dx

= (x2/2 + x)log⁡x – ∫1/x (x2/2 + x)dx

= (x2/2 + x)log⁡x – 1/2 ∫xdx – ∫dx

= (x + x2/2)log⁡x – 1/2 × x2/2 – x + c

Hence, I = (x + x2/2)log⁡x – 1/2 × x2/2 – x + c

Question 44. ∫ x2 tan-1xdx

Solution:

Given that, I = ∫ x2 tan-1⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = tan-1⁡x∫x2 dx – ∫(1/(1 + x2) ∫x2 dx) dx

= tan-1⁡x(x3/3) – 1/3∫x3/(1 + x2) dx

= 1/3 x3 tan-1⁡x – 1/3 ∫(x – x/(1 + x2))dx

= 1/3 x3 tan-1⁡x – 1/3 × x2/2 + 1/3 ∫x/(1 + x2) dx

Hence, I = 1/3 x3tan-1x – 1/6 x+ 1/6 log⁡|1 + x2| + c

Question 45. ∫(elogx + sin⁡x) cos⁡xdx

Solution:

Given that, I = ∫(elogx + sin⁡x)cos⁡xdx

= ∫(x + sin⁡x)cos⁡xdx

= ∫xcos⁡xdx + ∫sin⁡xcos⁡xdx

= [x∫cos⁡xdx – ∫(1]cos⁡xdx)dx] + 1/2 ∫sin⁡2xdx

= [xsin⁡x – ∫ sin⁡xdx] + 1/2 (-(cos⁡2x)/2) + c

I = xsin⁡x+cos⁡x – 1/4 cos⁡2x + c

= xsin⁡x + cos⁡x – 1/4 [1 – 2sin2⁡x] + c

= xsin⁡x + cos⁡x – 1/4 + 1/2 sin2x + c

= xsin⁡x + cos⁡x – 1/4 + 1/2 sin2x + c

Hence, I = xsin⁡x + cos⁡x + 1/2 sin2⁡x + d   [d = c-/4]

Question 46. ∫((xtan-1⁡x))/(1 + x2)3/2 dx

Solution:

Given that, I = ∫((xtan-1⁡x))/(1 + x2)3/2dx

Let us considered tan-1⁡x = t

1/(1 + x2) dx = dt

I = ∫(t tan⁡t)/√(1 + tan2⁡t) dt

= ∫(t × tan⁡t)/(sec⁡t) dt

= ∫t (sin⁡t)/(cos⁡t) cos⁡tdt

= ∫tsin⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = [t]sin⁡tdt – ∫(1)sin⁡tdt)dt]

= [-tcos⁡t + ∫cos⁡tdt]

= [-tcos⁡t + sin⁡t] + c

= -(tan-1⁡x)/√(1 + x2) + x/√(1 + x2) + c

Hence, I = -(tan-1⁡x)/√(1 + x2) + x/√(1 + x2) + c

Question 47. ∫ tan-1(√x)dx

Solution:

Given that, I = ∫ tan-1(√x)dx

Let us considered x = t2

dx = 2tdt

I = ∫2ttan-1tdt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get 

= 2[tan-1)⁡t∫tdt – ∫(1/(1 + t2) ∫tdt)dt]

= 2[t2/2 tan-1⁡t – ∫t2/2(1 + t2)dt]

= t2 tan-1⁡t – ∫(t2 + 1 – 1)/(1 + t2)dt

= t2 tan-1⁡t – ∫(1 – 1/(1 + t2))dt

= t2 tan-1t – t + tan-1⁡t + c

= (t2 + 1) tan-1⁡t – t + c

Hence, I = (x + 1)tan-1⁡√x – √x + c

Question 48. ∫x3 tan-1xdx

Solution:

Given that, I = ∫x3 tan-1xdx

= tan-1⁡x∫x3dx – (∫(dtan-1⁡x)/dx (∫x3 dx)dx)

= tan-1⁡x x4/4 – (∫1/(1 + x2) (x4/4)dx)

= tan-1⁡x x4/4 – (∫1/(1 + x2) (x4/4)dx) 

= tan-1⁡x x4/4 – (∫1/(1 + x2) (x4/4)dx)

∫ 1/(1 + x2) (x4/4)dx = 1/4 [∫1/(1 + x2) dx + (x2 – 1)dx]

∫ 1/(1 + x2) (x4/4)dx = 1/4 [tan-1⁡x + x3/3 – x]

Hence, I = x4/4 tan-1⁡x – 1/4 [tan-1⁡x + x3/3 – x] + c

Question 49. ∫xsin⁡xcos⁡2xdx

Solution:

Given that, I = ∫xsin⁡xcos⁡2xdx

= 1/2 ∫x(2sin⁡xcos⁡2x)dx

= 1/2 ∫x(sin⁡(x + 2x) – sin⁡(2x – x))dx

= 1/2 ∫x(sin⁡3x – sin⁡x)dx

= 1/2[x](sin⁡3x – sin⁡x)dx – ∫ (1)(sin⁡3x – sin⁡x)dx)dx]

= 1/2 [x((-cos⁡3x)/3 + cos⁡x) – ∫(-(cos⁡3x)/3 + cos⁡x)dx]

Hence, I = 1/2 [-x (cos⁡3x)/3 + xcos⁡x + 1/9 sin⁡3x – sin⁡x] + c

Question 50. ∫(tan-1x2)xdx

Solution:

Given that, I = ∫(tan-1⁡x2)xdx

Let us considered x2 = t

2xdx = dt

I = 1/2∫tan-1tdt

= 1/2∫1tan-1tdt

= 1/2 [tan-1⁡t∫dt – (∫1/(1 + t2)∫dt)dt]

= 1/2 [t × tan-1⁡t – ∫t/(1 + t2) dt]

= 1/2 t × tan-1⁡t – 1/4∫2t/(1 + t2) dt

= 1/2 t × tan-1⁡t – 1/4 log⁡|1 + t2| + c

Hence, I = 1/2 x2 tan-1⁡x2 – 1/4 log⁡|1 + x4| + c

Question 51. ∫xdx/√(1 – x2)

Solution:

Given that, I = ∫xdx/√(1 – x2)

Let first function be sin-1⁡x and second function be x/√(1 – x2).

Now, first we find the integral of the second function, 

∫xdx/√(1 – x2)

Now, put t = 1 – x2

Then dt = -2xdx

Therefore,

∫ xdx/√(1 – x2) = -1/2 ∫dt/√t = -√t = -√(1 – x2)

Hence,

∫(xsin-1x)/√(1 – x2) dx

= (sin-1⁡x)(-√(1 – x2) – ∫1/√(1 – x2) * (-√(1 – x2))dx

= -√(1 – x2) sin-1⁡x + x + c

= x – √(1 – x2) sin-1⁡x + c

Question 52. ∫sin3√x dx

Solution:

Given that, I = ∫sin3√x dx

Let us considered √x = t

x = t2

dx = 2tdt

I = 2∫ tsin3⁡tdt

= 2∫t((3sin⁡t – sin⁡3t)/4)dt

= 1/2 ∫t(3sin⁡t – sin⁡3t)dt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = 1/2 [t(-3cos⁡t + 1/3 cos⁡3t) – ∫(-3cos⁡t + (cos⁡3t)/3)dt]

= 1/2 [(-9tcos⁡t + tcos⁡3t)/3 – {-3sin⁡t + (sin⁡3t)/9}] + c

= 1/2 [(-9tcos⁡t + tcos⁡3t)/3 + (27sin⁡t – 3sin⁡3t)/9] + c

= 1/18[-27tcos⁡t + 3tcos⁡3t + 27sin⁡t – 3sin⁡3t] + c

Hence, I = 1/18[3√x cos⁡3√x + 27sin⁡√x – 27√x cos⁡√x – 3sin⁡3√x] + c

Question 53. ∫ xsin3xdx

Solution:

Given that, I = ∫ xsin3⁡xdx

= ∫x((3sin⁡x – sin⁡3x)/4)dx

= 1/4 ∫x(3sin⁡x – sin⁡3x)dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 1/4 [x∫ (3sin⁡x – sin⁡3x)dx – ∫(1)(3sin⁡x – sin⁡3x)dx)dx]

= 1/4 [x(-3cos⁡x + (cos⁡3x)/3) – ∫(-3cos⁡x + (cos⁡3x)/3)dx]

= 1/4 [-3xcos⁡x + (xcos⁡3x)/3 + 3sin⁡x – (sin⁡3x)/9] + c

Hence, I = 1/36[3xcos⁡3x – 27xcos⁡x + 27sin⁡x – sin⁡3x] + c

Question 54. ∫cos3√x dx

Solution:

Given that, I = ∫cos3√x dx

Let us considered x = t²

dx = 2tdt

= 2∫tcos3⁡tdt

= 2∫t((3cos⁡t + cos⁡3t)/4)dt

= 1/2 ∫t(3cos⁡t + cos⁡3t)dt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = 1/2 [t(3sin⁡t + 1/3 sin⁡3t) + ∫(1 × 3sin⁡t + (sin⁡3t)/3)dt]

= 1/2 [t((9sin⁡t + sin⁡3t)/3) + 3cos⁡t(cos⁡3t)/9] + c

= 1/18[27tsin⁡t + 3tsin⁡3t + 9cos⁡t + cos⁡3t] + c

Hence, I = 1/18[27√x sin⁡√x + 3√x sin⁡3√x + 9cos⁡√x + cos⁡3√x] + c

Question 55. ∫xcos3xdx

Solution:

Given that, I = ∫xcos3⁡xdx

= ∫x((3cos⁡x + cos⁡3x)/4)dx

= 1/4 ∫x(3cos⁡x + cos⁡3x)dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = 1/4 [x∫(3cos⁡x + cos⁡3x)dx – ∫(1)(3cos⁡x + cos⁡3x)dx)dx]

= 1/4 [x(3sin⁡x + (sin⁡3x)/3) – ∫ (3sin⁡x + (sin⁡3x)/3)dx]

= 1/4 [3xsin⁡x + (xsin⁡3x)/3 + 3cos⁡x + (cos⁡3x)/9] + c

Hence, I = (3xsin⁡x)/4 + (xsin⁡3x)/12 + (3cos⁡x)/4 + (cos⁡3x)/36 + c

Question 56. ∫tan-1√((1 – x)/(1 + x))

Solution:

Given that, I = ∫tan-1√((1 – x)/(1 + x))

Let us considered x = cos⁡θ

dx = -sin⁡θdθ

I = ∫ tan-1⁡(tan⁡θ/2)(-sin⁡θ)dθ

=-1/2 ∫θsin⁡θdθ

Let θ = u and sin⁡θdθ = v 

So that sin⁡θ = ∫vdθ

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = -1/2 (-θcos⁡θ – ∫-cos⁡θdθ)

= -1/2(-θcos⁡θ + sin⁡θ)+c

= -1/2 (-θcos⁡θ + √(1 – cos2⁡θ)) + c

= -1/2 (-xcos-1⁡x + √(1 – x2)) + c

Question 57. ∫sin-1√(x/(a + x)) dx

Solution:

Given that, I = ∫sin-1⁡√(x/(a + x)) dx

Let us considered x = atan2θ

dx = 2atan⁡θsec2⁡θdθ

I = ∫(sin-1⁡√((atan2⁡θ)/(a + atan2⁡θ))(2atan⁡θsec2θ)dθ

= ∫ (sin-1√((tan2θ)/(sec2θ)))(2atan⁡θsec2θ)dθ

= ∫ sin-1(sin⁡θ)(2atan⁡θsec2θ)dθ

= ∫ 2θatan⁡θsec2θdθ

= 2a∣θ(tan⁡θsec2⁡θ)dθ)

= ∫2θatan⁡θsec2θdθ

= 2a∫θ(tan⁡θsec2⁡θ)dθ

= 2a[θ]tan⁡θsec2θdθ – ∫(∫tan⁡θsec2⁡θdθ)dθ]

= 2a[θ (tan2⁡θ)/2 – ∫(tan2θ)/2 dθ]

= aθtan2θ – 2a/2∫(sec2θ – 1)dθ

= aθtan2θ – atan⁡θ + aθ + c

= a(tan-1⁡√(x/a)) x/a – a√(x/a) + atan-1⁡√(x/a) + c

Hence, I = xtan-1⁡√(x/a) – √ax + atan-1⁡√(x/a) + c

Question 58. ∫(x3 sin-1⁡x²)/√(1 – x4) dx

Solution:

Given that, I = ∫(x3 sin-1x²)/√(1 – x4) dx

Let us considered sin-1⁡x² = t

(1/√(1 – x4)(2x)dx = dt

I = ∫(x² sin-1⁡x²)/√(1 – x4) xdx

= ∫(sin⁡t)t dt/2

= 1/2∫tsin⁡tdt

= 1/2 [t∫sin⁡tdt – ∫(1∫sin⁡tdt)dt]

= 1/2 [t(-cost)dt – ∫(1∫(-cost))dt]

= 1/2[-tcost + sint] + c

Hence, I = 1/2 [x2 – √(1 – x4) sin(-1)⁡x2] + c

Question 59. ∫(x2 sin-1⁡x)/(1 – x2)3/2 dx

Solution:

Given that, I = ∫(x2 sin-1x)/(1 – x2)3/2dx

Let us considered sin-1⁡x = t

(1/√(1 – x2) dx = dt

I = ∫(sin2t × t)/((1 – sin2t)) dt

= ∫(tsin2t)/(cos2t) dt

= ∫t × tan2tdt

= ∫t(sec2⁡t – 1)dt

= ∫tsec2⁡tdt – t2/2 + c

= t∫sec2tdt – ∫(1∫sec2tdt)dt – t2/2 + c

= t × tan⁡t – ∫tan⁡tdt – t2/2 + c

= t × tan⁡t – log⁡sec⁡t – t2/2 + c

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

Question 60. ∫cos-1(1 – x2/ 1 + x2) dx

Solution:

Given that, I = ∫cos-1(1 – x2/ 1 + x2) dx

Let us considered, x = tant

dx = sec2tdt

I = ∫cos-1(1 – tan2t/ 1 + tan2t) sec2tdt

= ∫ 2t sec2tdt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = 2[t∫sec2tdt – ∫(1 ∫sec2tdt)dt]

= 2[t tan2t – ∫tant dt]

= 2[t tan2t – log sect] + c

= 2[x tan2x – log √1 + x2] + c

Hence, I = 2[xtan2x – log √1 + x2] + c

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