Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.2

In this exercise, we delve into the concept of integrals which is one of the fundamental aspects of calculus. The Integrals are used extensively in various fields such as physics, engineering, and economics for calculating areas, volumes, central points, and many other useful things. Exercise 7.2 in Chapter 7 focuses on the application of basic integration techniques allowing the students to strengthen their understanding and application of integrals in the solving mathematical problems.

Integrals

The Integrals are mathematical expressions that represent the area under a curve in a graph often referred to as the antiderivative. The process of integration is the inverse of differentiation and it is a crucial operation in the calculus. The Integrals can be definite or indefinite with the definite integrals providing a numerical value representing the area and indefinite integrals yielding a function as a result.

In this article, we will find the NCERT solutions to all the questions of Exercise 7.2 of Chapter 7 Integrals, Class 12.

Question 1: \frac{2x}{1+x^2}

Answer

Let 1+x² = t

⇒ 2x dx = dt

⇒ \int \frac{2x}{1+x^2} dx

=\int \frac{1}{t} dt

= log|t| + C

= log|1+x²|+C

Question 2:\frac{(logx)^2}{x}

Answer

Let log |x| = t

⇒ \frac{1}{x} dx = dt

⇒\int \frac{(log|x|)^2}{x} dx = \int t^2 dt

= \frac{t^3}{3} + C

= \frac{(log|x|)^3}{3} + C

Question 3:\frac{1}{x+xlogx}

Answer:

\frac{1}{x+xlogx} = \frac{1}{x(1+logx)}

Let 1+log x = t

\frac{1}{x} dx = dt

=>\int \frac{1}{x(1+logx)} dx = \int \frac{1}{t} dt

= log |t| + C

= log|1+logx| + C

Question 4: sinx.sin(cosx)

Answer:

sinx.sin(cosx)

Let cosx = t

-sinx dx = dt

\int sinx.sin(cosx) d x = -\int sint \,dt

=-[-cost]+C

= cost + C

= cos(cosx) + C

Question 5: sin(ax+b) cos(ax+b)

Answer:

sin(ax+b) cos(ax+b) = \frac{2sin(ax+b) cos(ax+b)}{2}

\frac{sin2(ax+b)}{2}

Let 2(ax+b) = t

So 2a dx = dt

=> \int \frac{sin2(ax+b)}{2} dx

=\frac{1}{2} \int \frac{sint}{2a} dt

=\frac{1}{4a} [-cost] + C

==\frac{-1}{4a} cos2(ax+b) + C

Question 6:\sqrt{ax+b}

Answer:

Let ax+b = t

a dx = dt

dx = \frac{1}{a} dt

=> \int \sqrt{ax+b} dx = \frac{1}{a} \int t^{\frac{1}{2}} dt

= \frac{1}{a}\{\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\}+C

= \frac{2}{3a}(ax+b)^{\frac{3}{2}} + C

Question 7:x\sqrt{x+2}

Answer:

Let (x+2) = t

dx = dt

=> \int x\sqrt{x+2} \, dx = \int (t-2)\sqrt{t} \, dt

=\int\{t^{\frac{3}{2}}-2t^{\frac{1}{2}}\}dt

=\int t^{\frac{3}{2}} dt- \int t^{\frac{1}{2}} dt

= \frac{t^{\frac{5}{2}}}{\frac{5}{2}}-2\{\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\}+C

=\frac{2}{5}t^{\frac{5}{2}}-\frac{4}{3}t^{\frac{3}{2}} + C

=\frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{4}{3}(x+2)^{\frac{3}{2}} + C

Question 8:x\sqrt{1+2x^2}

Answer:

Let 1+2x² = t

4x dx = dt

=> \int x\sqrt{1+2x^2} dx = \int \frac{\sqrt{t}}{4} dt

= \frac{1}{4} \int t^{\frac{1}{2}} dt

= \frac{1}{4}\{\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\} +C

= \frac{1}{6}(1+2x^2)^{\frac{3}{2}} + C

Question 9: (4x+2)\sqrt{x^2+x+1}

Answer:

Let x² + x + 1 = t

(2x+1) dx = dt

\int (4x+2)\sqrt{x^2+x+1}dx

=\int 2\sqrt{t} dt =2\int \sqrt{t} dt

= 2\frac{t^{\frac{3}{2}}}{\frac{3}{2}} + C

= \frac{4}{3}(x^2+x+1)^{\frac{3}{2}} + C

Question 10: \frac{1}{x-\sqrt{x}}

Answer:

\frac{1}{x-\sqrt{x}} = \frac{1}{\sqrt{x}(\sqrt{x}-1)}

Let (\sqrt{x}-1) = t

\frac{1}{2\sqrt{x}} dx = dt

=>\int \frac{1}{\sqrt{x}(\sqrt{x}-1)}dx = \int \frac{2}{t} dt

= 2 log |t| + C

2log|\sqrt{x}-1| + C

Question 11: \frac{x}{\sqrt{x+4}},x>0

Answer:

Let x+4 = t

dx = dt

\int \frac{x}{\sqrt{x+4}} dx = \int \frac{(t-4)}{\sqrt{t}} dt

= \int (\sqrt{t}-\frac{4}{\sqrt{t}})dt

=\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4\{\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\}+C

=\frac{2}{3}t^{\frac{3}{2}} - 8t^{\frac{1}{2}}+ C

=\frac{2}{3}t.t^{\frac{1}{2}} - 8t^{\frac{1}{2}}+ C

=\frac{2}{3}(x+4)^{\frac{1}{2}} (x+4-12)+ C

=\frac{2}{3} \sqrt{x+4}(x-8) + C

Question 12: (x^3-1)^{\frac{1}{3}}x^5

Answer:

Let x^3-1 = t

3x^2 dx= dt

=> \int (x^3-1)^{\frac{1}{3}}x^5 dx = \int(x^3-1)^{\frac{1}{3}}x^3.x^2 dx

=\int t^{\frac{1}{3}}(t+1)\frac{dt}{3}

=\frac{1}{3}\int\{t^{\frac{4}{3}}+t^{\frac{1}{3}}\}dt

=\frac{1}{3}\{\frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\}+C

= \frac{1}{3}[ \,\frac{3}{7}t^{\frac{7}{3}}+\frac{3}{4}t^{\frac{4}{3}}] \,+C

= \,\frac{1}{7}(x^3-1)^{\frac{7}{3}}+\frac{3}{4}(x^3-1)^{\frac{4}{3}} \,+C

Question 13:\frac{x^2}{(2+3x^3)^3}

Answer:

Let 2+3x^3=t

9x^2 dx=dt

\int \frac{x^2}{(2+3x^3)^3}dx = \frac{1}{9}\int \frac{dt}{t^3}

=\frac{1}{9}[ \,\frac{t^{-2}}{-2}] \,+C

=\frac{-1}{18}[ \,\frac{1}{t^{2}}] \,+C

=\frac{-1}{18(2+3x^3)^2}+C

Question 14:\frac{1}{x(logx)^m},x>0

Answer:

Let log x = t

1/x dx = dt

=> \int \frac{1}{x(logx)^m}dx = \int \frac{dt}{(t)^m}

=\{\frac{t^{-m+1}}{1-m}\}+C

=\{\frac{(logx)^{-m+1}}{1-m}\}+C

Question 15:\frac{x}{9-4x^2}

Answer:

Let 9-4x^2 = t

-8x dx = dt

=>\int \frac{x}{9-4x^2} dx=\frac{-1}{8}\int \frac{1}{t}dt

=\frac{-1}{8} log|t| +C

=\frac{-1}{8} log|9-4x^2|+C

Question 16:e^{2x+3}

Answer:

Let 2x+3 = t

2dx = dt

=> \int e^{2x+3} dx = \frac{1}{2}\int e^t dt

=\frac{1}{2}(e^t)+C

=\frac{1}{2}e^{2x+3}+C

Question 17:\frac{x}{e^{x^2}}

Answer:

Let x^2 = t

2x dx = dt

=> \int \frac{x}{e^{x^2}} dx = \frac{1}{2} \int \frac{1}{e^t}dt

=\frac{1}{2}[ \,\frac{e^{-t}}{-1}] \,+C

= \frac{-1}{2}e^{-x^2}+C

=\frac{-1}{2e^{x^2}}+C

Question 18:\frac{e^{tan^{-1}x}}{1+x^2}

Answer:

Let tan^-1 x =t

\frac{1}{1+x^2} dx = dt

=> \int \frac{e^{tan^{-1}x}}{1+x^2} dx = \int e^t dt

=e^t+C

=e^{tan^{-1}x}+C

Question 19:\frac{e^{2x}-1}{e^{2x}+1}

Answer:

\frac{e^{2x}-1}{e^{2x}+1}

Dividing numerator and denominator by e^x , we obtain

\frac{\frac{(e^{2x}-1)}{e^x}}{\frac{(e^{2x}+1)}{e^x}}

\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}

Let\,\, {e^{x}+e^{-x}}=t

( {e^{x}-e^{-x}})dx=dt

\int \frac{e^{2x}-1}{e^{2x}+1} dx=\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} dx=\int \frac{dt}{t}

=log|t|+C

=log|{e^{x}+e^{-x}}|+C

Question 20:\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}

Answer:

Let Let\,\, {e^{2x}+e^{-2x}}=t

(2\,\, {e^{2x}-2e^{-2x}})dx=dt

2(\,\, {e^{2x}-e^{-2x}})dx=dt

=> \int \{\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\} dx = \int \frac{dt}{2t}

=\frac{1}{2}\int{\frac{1}{t}}dt

=\frac{1}{2}log|t|+C

=\frac{1}{2}log|\,\, {e^{2x}+e^{-2x}}|+C

Question 21:tan^2(2x-3)

Answer:

tan^2(2x-3) = sec^2(2x-3)-1

Let \, 2x-3 = t

2 dx = dt

\int tan^2(2x-3) dx= \int[ \,sec^2(2x-3)-1] \,dx

=\frac{1}{2}\int (sec^2t)dt-\int 1 dt

=\frac{1}{2}tant-t+C

=\frac{1}{2}tan(2x-3)-(2x - 3)+C

Question 22:sec^2(7-4x)

Answer:

Let 7-4x = t

-4dx = dt

\int sec^2(7-4x) dx = \frac{-1}{4}\int sec^2t dt

=\frac{-1}{4}(tant)+C

=\frac{-1}{4}tan(7-4x)+C

Question 23: \frac{sin^{-1}x}{\sqrt{1-x^2}}

Answer:

Let\,\,sin^{-1}x = t

\frac{1}{\sqrt{1-x^2}}dx=dt

\int \frac{sin^{-1}x}{\sqrt{1-x^2}}dx=\int t dt

=\frac{t^2}{2}+C

=\frac{(sin^{-1}x )^2}{2}+C

Question 24:\frac{2cosx-3sinx}{6cosx+4sinx}

Answer:

\frac{2cosx-3sinx}{6cosx+4sinx}=\frac{2cosx-3sinx}{2(3cosx+2sinx)}

Let 3cosx+2sinx=t

(-3 sinnx+2 cosx)dx = dt

\int \frac{2cosx-3sinx}{6cosx+4sinx}dx = \int\frac{dt}{2t}

=\frac{1}{2}\int \frac{1}{t}dt

=\frac{1}{2}log|t|+C

=\frac{1}{2}log|2sinx+3cosx|+C

Question 25:\frac{1}{cos^2x(1-tanx)^2}

Answer:

Let (1-tanx)=t

-sec^2x dx=dt

=> \int \frac{sec^2x}{(1-tanx)^2}dx=\int \frac{-dt}{t^2}

=-\int t^{-2} dt

=+\frac{1}{t}+C

=\frac{1}{(1-tanx)}+C

Question 26:\frac{cos\sqrt{x}}{\sqrt{x}}

Answer:

Let \sqrt{x}=t

\frac{1}{2\sqrt{x}}dx=dt

\int \frac{cos\sqrt{x}}{\sqrt{x}} dx=2\int cost dt

=2sint+C=>2sin\sqrt{x}+C

Question 27:\sqrt{sin2x}cos2x

Answer:

Let sin2x=t

2 cos2x dx = dt

=>\int \sqrt{sin2x}cos2x\,dx = \frac{1}{2}\int \sqrt{t}\,dt

=\frac{1}{2}\{\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\}+C

=\frac{1}{3}t^{\frac{3}{2}}+C

=\frac{1}{3}(sin2x)^{\frac{3}{2}}+C

Question 28:\frac{cosx}{\sqrt{1+sinx}}

Answer:

Let 1+sinx = t

cosx dx = dt

=> \int \frac{cosx}{\sqrt{1+sinx}} dx = \int \frac{dt}{\sqrt{t}}

=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C

=2\sqrt{t}+C

=2\sqrt{1+sinx}+C

Question 29:cotx \,\,log \,sinx

Answer:

Let log\,sinx = t

=>\frac{1}{sinx}.cosx \, dx=dt

cotx \,dx=dt

=>\int cotx \,\,log \,sinx dx = \int t\, dt

=\frac{t^2}{2}+C

=\frac{(log\,sinx)^2}{2}+C

Question 30:\frac{sinx}{1+cosx}

Answer:

Let 1+cosx = t

-sinx \,dx = dt

=>\int \frac{sinx}{1+cosx} dx=\int -\frac{dt}{t}

=-log|t|+C

=-log|1+cosx|+C

Question 31:\frac{(sinx)}{(1+cosx)^2}

Answer:

Let 1+cosx = t

-sinx dx = dt

=> \int \frac{(sinx)}{(1+cosx)^2} dx=\int -\frac{dt}{t^2}

=-\int t^{-2}dt

=\frac{1}{t}+C

=\frac{1}{1+cosx}+C

Question 32: \frac{1}{1+cot\,x}

Answer:

Let I = \int \frac{1}{1+cot\,x} dx

=\int \frac{1}{1+\frac{cosx}{sinx}}dx

=\int \frac{sinx}{sinx+cosx} dx

=\frac{1}{2}\int \frac{2sinx}{sinx+cosx} dx

=\frac{1}{2}\int \frac{(sinx+cosx)+(sinx-cosx)}{sinx+cosx} dx

=\frac{1}{2}\int 1\,dx + \frac{1}{2}\int \frac{(sinx-cosx)}{sinx+cosx} dx

=\frac{1}{2}x + \frac{1}{2}\int \frac{(sinx-cosx)}{sinx+cosx} dx

Let sinx+cosx = t => (cosx-sinx)dx=dt

So, I=\frac{1}{2}x+\frac{1}{2}\int\frac{-dt}{t}

=\frac{1}{2}x-\frac{1}{2}log|t|+C

=\frac{x}{2}-\frac{1}{2}log|sinx+cosx|+C

Question 33:\frac{1}{1-tanx}

Answer:

Let I=\frac{1}{1-tanx}

=\int \frac{1}{1-\frac{sinx}{cosx}}dx

=\int \frac{cosx}{cosx-sinx} dx

=\frac{1}{2}\int \frac{2cosx}{cosx-sinx} dx

=\frac{1}{2}\int \frac{(cosx-sinx)+ (sinx+cosx)}{cosx-sinx} dx

=\frac{1}{2}\int 1\,dx + \frac{1}{2}\int \frac{(sinx+cosx)}{cosx-sinx} dx

=\frac{1}{2}x + \frac{1}{2}\int \frac{(cosx+sinx)}{cosx-sinx} dx

Let cosx - sinx = t => (-sinx-cosx)dx = dt

So, I=\frac{1}{2}x+\frac{1}{2}\int\frac{-dt}{t}

=\frac{1}{2}x-\frac{1}{2}log|t|+C

=\frac{x}{2}-\frac{1}{2}log|cosx-sinx|+C

Question 34:\frac{\sqrt{tanx}}{sinx\,cosx}

Answer:

Let\,I = \int \frac{\sqrt{tanx}}{sinx\,cosx} dx

\int \frac{\sqrt{tanx}\,*cosx}{sinx\,cosx*\,cosx} dx

=\int \frac{\sqrt{tanx}}{tanx\,cos^2x} dx

=\int \frac{sec^2}{\sqrt{tanx}}dx

Let\,tanx=t=>sec^2x\,dx=dt

I=\int \frac{dt}{\sqrt{t}}

=2\sqrt{t}+C

=2\sqrt{tanx}+C

Question 35:\frac{(1+logx)^2}{x}

Answer:

Let 1+ log x = t

\frac{1}{x}dx = dt

\int \frac{(1+logx)^2}{x} dx = \int t^2 dt

=\frac{t^3}{3}+C

=\frac{(1+logx)^3}{3}+C

Question 36:\frac{(x+1)(x+logx)^2}{x}

Answer:

\frac{(x+1)(x+logx)^2}{x}=\frac{x+1}{x}(x+logx)^2=(1+\frac{1}{x})(x+logx)^2

Let \,(x+logx)=t

(1+\frac{1}{x})dx=dt

\int (1+\frac{1}{x})(x+logx)^2dx=\int t^2dt

=\frac{t^3}{3}+C

=\frac{1}{3}(x+logx)^3+C

Question 37:\frac{x^3sin(tan^{-1}x^4)}{1+x^8}

Answer:

Let\,x^4=t=>4x^3dx=dt

=>\int \frac{x^3sin(tan^{-1}x^4)}{1+x^8}dx=\frac{1}{4}\int\frac{sin(tan^{-1}t)}{1+t^2} ------(i)

Let\,tan^{-1}t=u

\frac{1}{1+t^2}dt=du

From (i),we obtain

\int \frac{x^3sin(tan^{-1}x^4)}{1+x^8} dx=\frac{1}{4}\int sinu\,du

=\frac{1}{4}(-cosu)+C

=\frac{-1}{4}cos(tan^{-1}t)+C

=\frac{-1}{4}cos(tan^{-1}x^4)+C

Choose the correct answer In Question 38 and 39

Question 38:\int \frac{10x^9+10^xlog }{x^{10}+{10}^{x}} dx

(A)\,\,10^x-x^{10}+C

(B)\,\, 10^x+x^{10}+C

(c) \,\,(10^x-x^{10})^{-1}+C

(D)\,\,log(10^x+x^{10})+C

Answer:

Let x¹⁰ + 10^x = t

Hence, (10x⁹ + 10^x log_e10 )dx = dt

\int \frac {10x^9+ 10^xlog_e10} {x^10 + 10^x} dx = \int\frac{dt}{t}

= log t + C

= lod(10^x + x^10) + C

Hence, Correct Answer is (D)\,\,log(10^x+x^{10})+C

Question 39:\int \frac{dx}{sin^2x\, cos^2x} \,\,equals

(A)\, \, tanx+cotx+C

(B)\, \, tanx-cotx+C

(C)\, \, tanx\,\,cotx+C

(D)\, \, tanx-cot2x+C

Answer:

Let us take I = \int \frac{dx}{sin^2x\, cos^2x}

We know sin²x + cos²x = 1

Hence the above equation becomes

\int \frac{sin^2x + cos^2x}{sin^2x\, cos^2x} dx

=\int \frac{sin^2x}{sin^2x\, cos^2x} dx + \int \frac{ cos^2x}{sin^2x\, cos^2x} dx

=\int {sec^2x} dx + \int {cosec^2x} dx

= tan x - cot x +C

Hence, Correct Answer is (B)\, \, tanx-cotx+C

Also Check:

• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.1
• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.3
• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.4
• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.5
• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.6
• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.7

Conclusion

Exercise 7.2 provides a solid foundation for the mastering the techniques of integration. The problems included are designed to help we understand how to apply the fundamental concepts of integrals in the various contexts. By practicing these problems we'll be well-prepared to tackle more advanced topics in the calculus and apply these mathematical principles to the real-world scenarios.