Question
Integrate the function: $\frac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)}}{{1 + {x^8}}}$
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Answer
Let $I = \int {\frac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)}}{{1 + {x^8}}}dx} $
$ = \frac{1}{4}\int {\sin \left( {{{\tan }^{ - 1}}{x^4}} \right).\frac{{4{x^3}}}{{1 + {x^8}}}dx} $…(i)
Putting $\tan^{-1}x^4 = t$
$\Rightarrow \frac{1}{{1 + {{\left( {{x^4}} \right)}^2}}}\frac{d}{{dx}}{x^4} = \frac{{dt}}{{dx}}$
$\Rightarrow \frac{{4{x^3}}}{{1 + {x^8}}}dx = dt$
From eq. (i), $I = \frac{1}{4}\int {\sin tdt} $
$= \frac{{ - 1}}{4}\cos t + c$
$ = \frac{{ - 1}}{4}\cos \left( {{{\tan }^{ - 1}}{x^4}} \right) + c$
$ = \frac{1}{4}\int {\sin \left( {{{\tan }^{ - 1}}{x^4}} \right).\frac{{4{x^3}}}{{1 + {x^8}}}dx} $…(i)
Putting $\tan^{-1}x^4 = t$
$\Rightarrow \frac{1}{{1 + {{\left( {{x^4}} \right)}^2}}}\frac{d}{{dx}}{x^4} = \frac{{dt}}{{dx}}$
$\Rightarrow \frac{{4{x^3}}}{{1 + {x^8}}}dx = dt$
From eq. (i), $I = \frac{1}{4}\int {\sin tdt} $$= \frac{{ - 1}}{4}\cos t + c$
$ = \frac{{ - 1}}{4}\cos \left( {{{\tan }^{ - 1}}{x^4}} \right) + c$
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