Q. $\int_{}^{}\frac{x^{3}\sin{\lbrack\tan^{- 1}{(x^{4})}\rbrack}}{1 + x^{8}}dx$ is equal to
A) $\frac{1}{4}\cos{\left\lbrack \tan^{- 1}\left( x^{4} \right) \right\rbrack + c}$
B) $\frac{1}{4}\sin{\left\lbrack \tan^{- 1}\left( x^{4} \right) \right\rbrack + c}$
C) $- \frac{1}{4}\cos{\left\lbrack \tan^{- 1}\left( x^{4} \right) \right\rbrack + c}$
D) $\frac{1}{4}\sec^{- 1}{\left\lbrack \tan^{- 1}\left( x^{4} \right) \right\rbrack + c}$
Explanation
(c)
Let $I = \int_{}^{}\frac{x^{3}\sin{\left\lbrack \tan^{- 1}\left( x^{4} \right) \right\rbrack dx}}{1 + x^{8}}$
Put $x^{4} = t\ \Rightarrow \ 4x^{3}\ dx = dt$
$\therefore\ \ I = \int_{}^{}{\frac{1}{4}.\ \ \frac{\sin{\lbrack\tan^{- 1}{(t)}\rbrack dt}}{1 + t^{2}}}$
Again, put $\tan^{- 1}{t = u}\ \ \Rightarrow \ \frac{1}{1 + t^{2}}dt = du$
$\therefore\ I = \int_{}^{}{\frac{1}{4}\sin u}du = - \frac{1}{4}\cos u + c = - \frac{1}{4}\cos{\left\lbrack \tan^{- 1}\left( x^{4} \right) \right\rbrack + c}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.