Question

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

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