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$\int \frac{x^{3} \sin \left(\tan ^{-1}\left(x^{4}\right)\right)}{1+x^{8}} d x$ is equal to
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Verified Answer
The correct answer is:
$\frac{-\cos \left(\tan ^{-1}\left(x^{4}\right)\right)}{4}+C$
Let $I=\int \frac{x^{3} \sin \left[\tan ^{-1}\left(x^{4}\right)\right]}{1+x^{8}}$
Let $\tan ^{-1}\left(x^{4}\right)=t$.
On differentiating w.r.t. $t$, we get
$\begin{gathered}
\frac{1}{1+x^{8}} \times 4 x^{3} d x=d t \\
\Rightarrow \quad \frac{x^{3}}{1+x^{8}} d x=\frac{1}{4} d t \\
I=\frac{1}{4} \int \sin t d t
\end{gathered}$
$=-\frac{1}{4} \cos t+C$
$=-\frac{1}{4} \cos \left(\tan ^{-1}\left(x^{4}\right)\right)+C$
Let $\tan ^{-1}\left(x^{4}\right)=t$.
On differentiating w.r.t. $t$, we get
$\begin{gathered}
\frac{1}{1+x^{8}} \times 4 x^{3} d x=d t \\
\Rightarrow \quad \frac{x^{3}}{1+x^{8}} d x=\frac{1}{4} d t \\
I=\frac{1}{4} \int \sin t d t
\end{gathered}$
$=-\frac{1}{4} \cos t+C$
$=-\frac{1}{4} \cos \left(\tan ^{-1}\left(x^{4}\right)\right)+C$
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