Question

$\int \frac{x^3\sin \left({\tan }^{-1}\left(x^4\right)\right)}{1+x^8}dx$ is equal to

• A. $\frac{-\cos \left({\tan }^{-1}\left(x^4\right)\right)}{4}+C$
• B. $\frac{\cos \left({\tan }^{-1}\left(x^4\right)\right)}{4}+C$
• C. $\frac{-\cos \left({\tan }^{-1}\left(x^3\right)\right)}{3}+C$
• D. $\frac{\sin \left({\tan }^{-1}\left(x^4\right)\right)}{4}+C$

Answer 1

Answer: (A)

${\text{Tan}}^{-1}{x}^4=t;\frac{4x^3}{1+x^8}dx=dt$
$I=\frac{1}{4}\int \sin tdt=\frac{-1}{4}\cos t+c$
$=\frac{-1}{4}\cos \left({\text{Tan}}^{-1}{x}^4\right)+c$