Get Answers to all your Questions

header-bg qa

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 121 Maths Textbook Solution.

Answers (1)


Answer:\frac{e^{m \tan ^{-1} x}}{m^{2}+1}\left(m \cos \left(\tan ^{-1} x\right)+\sin \left(\tan ^{-1} x\right)+C\right.

Given: \int \frac{e^{m \operatorname{tan}^{-1} x}}{\left(1+x^{2}\right)^{3 / 2}} d x

Hint: using integration by parts


\text { Let } I=\int \frac{e^{m \tan ^{-1} x}}{\left(1+x^{2}\right)^{3 / 2}} d x

I=\int \frac{e^{m \theta}}{\left(1+\tan ^{2} \theta\right)^{3 / 2}} \cdot \sec ^{2} \theta d \theta                             x=\tan \theta \cdot \theta=\tan ^{-1} x

                                                                                   Put  d x=\sec ^{2} \theta d \theta

I=\int \frac{e^{m \theta}}{\left(\sec ^{2} \theta\right)^{3 / 2}} \cdot \sec ^{2} \theta d \theta

I=\int \frac{e^{m \theta}}{\sec ^{3} \theta} \sec ^{2} \theta d \theta

I=\int \frac{e^{m \theta} d \theta}{\sec \theta}=\int e^{m \theta} \cos \theta d \theta\left[\because \frac{1}{\cos \theta}=\sec \theta\right]

I=\cos \theta \frac{e^{m \theta}}{m}-\int(-\sin \theta) \frac{e^{m \theta}}{m} d \theta \ldots u s i n g \text { Byparts }

I=\frac{1}{m} \cos \theta e^{m \theta}+\frac{1}{m} \int \sin \theta e^{m \theta} d \theta

I=\frac{1}{m} \cos \theta e^{m \theta}+\frac{1}{m}\left[\sin \theta \frac{e^{m \theta}}{m}-\int \cos \theta \frac{e^{m \theta}}{m} d \theta\right]

I=\frac{1}{m} \cos \theta e^{m \theta}+\frac{1}{m^{2}} \sin \theta e^{m \theta}-\frac{1}{m^{2}} \int \cos \theta e^{m \theta} d \theta

I=\frac{1}{m} \cos \theta \cdot e^{m \theta}+\frac{1}{m^{2}} \int \sin \theta e^{m \theta}-\frac{1}{m^{2}} I \ldots \text { w?ere } I=\int \cos \theta \cdot e^{m \theta} d \theta

\left(1+\frac{1}{m^{2}}\right) I=\frac{1}{m} \cos \theta \cdot e^{m \theta}+\frac{1}{m} \sin \theta e^{m \theta}+C

\left(\frac{m^{2}+1}{m^{2}}\right) I=e^{m \theta}\left(\frac{\cos \theta}{m}+\frac{\sin \theta}{m^{2}}\right)+C

\frac{\left(m^{2}+1\right)}{m^{2}} I=e^{m \theta}\left(\frac{m \cos \theta+\sin \theta}{m^{2}}\right)+m^{2} C

\left(m^{2}+1\right) I=e^{m \theta}(m \cos \theta+\sin \theta)+m^{2} C

I=\frac{e^{m \theta}}{m^{2}+1}(m \cos \theta+\sin \theta)+m^{2} C

=\frac{e^{m \tan ^{-1} x}}{m^{2}+1}\left(m \cos \left(\tan ^{-1} x\right)+\sin \left(\tan ^{-1} x\right)+C\right.

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support