Question

$\int \sin \left({\tan }^{-1}x\right)dx=$

• A. $\frac{1}{\sqrt{1+x^2}}+c$
• B. $\frac{1}{\sqrt{1-x^2}}+c$
• C. $\sqrt{1+x^2}+c$
• D. $\sqrt{1-x^2}+c$

Answer 1

Answer: (C)

$\sqrt{1+x^2}+c$

Let $I=\int \sin \left({\tan }^{-1}x\right)dx$

Take ${\tan }^{-1}x=\theta$

$x=\tan \theta$ $\Rightarrow dx={\sec }^2\theta d\theta$

Then,

$I=\int \sin \theta {\sec }^2\theta d\theta$

$=\int \tan \theta \sec \theta d\theta$

$=\sec \theta +c$

Now, ${\sec }^2\theta -{\tan }^2\theta =1$

i.e. ${\sec }^2\theta =1+x^2$

i.e. $\sec \theta =\sqrt{1+x^2}$

Then, $I=\sec \theta +c=\sqrt{1+x^2}+c$

Hence, $\int \sin \left({\tan }^{-1}x\right)dx=\sqrt{1+x^2}+c$