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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

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

Similar Questions

$\int \frac{dx}{\sqrt{2x - x^2}} = $

$\int \frac{1 + \tan x}{x + \log \sec x} \, dx = $

$\int \frac{dx}{x^2(x^4+1)^{3/4}}$ equals

If $\int \frac{\sin ^3 x}{\left(\cos ^4 x+3 \cos ^2 x+1\right) \tan ^{-1}(\sec x+\cos x)} d x=f(x)+C$,then $e^{f(x)}=$

Integrate the function $\frac{(1+\log x)^{2}}{x}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module