KCET · Maths · Indefinite Integration
\(\int \frac{x^{3} \sin \left(\tan ^{-1}\left(x^{4}\right)\right)}{1+x^{8}} d x\) 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 & Solution
Correct Answer
(A) \(\frac{-\cos \left(\tan ^{-1}\left(x^{4}\right)\right)}{4}+C\)
Step-by-step Solution
Detailed explanation
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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