Question

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

Answer 1

Solution : -

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

Let $ta{n}^{-1}{x}^4=t$

$\Rightarrow \frac{4x^3}{1+x^8}dx=dt$

$\Rightarrow \frac{x^3}{1+x^8}dx=\frac{dt}{4}$

$\Rightarrow I=\int \frac{sin\left(t\right).dt}{4}$

$=\frac{-1}{4}\left[cost\right]+c$

$=\frac{-cost}{4}+c$

Now $t=ta{n}^{-1}{x}^4$

$\therefore I=\frac{-cos\left(ta{n}^{-1}{x}^4\right)}{4}+c$