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Q37  Integrate the functions  \frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }

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Given function  \frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }

Assume that x^4 =t

\therefore 4x^3 dx =dt

\Rightarrow \int \frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }dx = \frac{1}{4} \int \frac{\sin(\tan^{-1} t)}{1+t^2}dt            ......................(1)

Now to solve further we take \tan ^{-1} t = u

\therefore \frac{1}{1+t^2} dt =du

So, from the equation (1), we will get

\Rightarrow \int \frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }dx =\frac{1}{4}\int \sin u\ du

= \frac{1}{4}(-\cos u) +C

Now back substitute the value of u,

= \frac{-1}{4}\cos (\tan^{-1} t) +C

and then back substituting the value of t,

= \frac{-1}{4}\cos (\tan^{-1} x^4) +C

Posted by

Divya Prakash Singh

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