int frac{x^{3} sin left(tan ^{-1}left(x^{4}ri | vedclass

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(A) Let $I = \int \frac{x^{3} \sin \left(	an ^{-1}\left(x^{4}ight)ight)}{1+x^{8}} dx$. Substitute $t = 	an ^{-1}\left(x^{4}ight)$. Differentia

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$\frac{-\cos \left(	an ^{-1}\left(x^{4}ight)ight)}{4}+C$

Vedclass · Answer

(A) Let $I = \int \frac{x^{3} \sin \left(	an ^{-1}\left(x^{4}ight)ight)}{1+x^{8}} dx$. Substitute $t = 	an ^{-1}\left(x^{4}ight)$. Differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{1+(x^{4})^{2}} \cdot 4x^{3} = \frac{4x^{3}}{1+x^{8}}$. Thus,$\frac{x^{3}}{1+x^{8}} dx = \frac{1}{4} dt$. Substituting these into the integral,we get $I = \int \sin(t) \cdot \frac{1}{4} dt$. $I = \frac{1}{4} \int \sin(t) dt = \frac{1}{4} (-\cos(t)) + C$. Substituting back $t = 	an ^{-1}\left(x^{4}ight)$,we get $I = -\frac{1}{4} \cos \left(	an ^{-1}\left(x^{4}ight)ight) + C$.

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

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