Question

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

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

Answer 1

The correct option is C $-\frac{\cos ({\tan }^{-1}{x}^4)}{4}+C$

$\int \frac{x^3\sin ({\tan }^{-1}{x}^4)}{1+x^8}dx$
Let
${\tan }^{-1}{x}^4=t\Rightarrow 4x^3\frac{1}{1+(x^4{)}^2}dx=dt$
Therefore
$\int \sin t\cdot \frac{dt}{4}=-\frac{\cos t}{4}+C=-\frac{\cos ({\tan }^{-1}{x}^4)}{4}+C$