Q107 / 180 · KCET 2021 · Mathematics

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

  • Correct answer
    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
    $\frac{\sin \left(\tan ^{-1}\left(x^{4}\right)\right)}{4}+C$

Solution

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 $\begin{gathered} \frac{1}{1+x^{8}} \times 4 x^{3} d x=d t \\ \Rightarrow \quad \frac{x^{3}}{1+x^{8}} d x=\frac{1}{4} d t \\ I=\frac{1}{4} \int \sin t d t \end{gathered}$ $=-\frac{1}{4} \cos t+C$ $=-\frac{1}{4} \cos \left(\tan ^{-1}\left(x^{4}\right)\right)+C$

Hence, the correct answer is option: 'A'.

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