int frac{x^3}{sqrt{1 - x^8}} , dx = | vedclass

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

Question

(C) Let $I = \int \frac{x^3}{\sqrt{1 - x^8}} \, dx$. Rewrite the integral as $I = \int \frac{x^3}{\sqrt{1 - (x^4)^2}} \, dx$. Substitute $t = x^4$,the

Vedclass · Accepted Answer

$\frac{1}{4} \sin^{-1}(x^4) + c$

Vedclass · Answer

(C) Let $I = \int \frac{x^3}{\sqrt{1 - x^8}} \, dx$. Rewrite the integral as $I = \int \frac{x^3}{\sqrt{1 - (x^4)^2}} \, dx$. Substitute $t = x^4$,then $dt = 4x^3 \, dx$,which implies $x^3 \, dx = \frac{1}{4} \, dt$. Substituting these into the integral,we get $I = \int \frac{1}{\sqrt{1 - t^2}} \cdot \frac{1}{4} \, dt$. $I = \frac{1}{4} \int \frac{1}{\sqrt{1 - t^2}} \, dt$. Using the standard integral $\int \frac{1}{\sqrt{1 - t^2}} \, dt = \sin^{-1}(t) + c$,we get $I = \frac{1}{4} \sin^{-1}(t) + c$. Substituting back $t = x^4$,we obtain $I = \frac{1}{4} \sin^{-1}(x^4) + c$.

$\int \frac{x^3}{\sqrt{1 - x^8}} \, dx = $

Similar Questions

$\int \frac{3x^2}{\sqrt{9 - 16x^6}} \, dx = $

$\int \frac{e^x(x + 1)}{\cos^2(x e^x)} dx = $

$\int x^2 \sec(x^3) \, dx = $

If $\int \frac{x^3}{\sqrt{1+x^2}} d x=A(1+x^2)^{\frac{3}{2}}+B(1+x^2)^{\frac{1}{2}}+C$,then $A+B=$

$\int x \operatorname{Tan}^{-1} \sqrt{\frac{1+x^2}{1-x^2}} \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module