If int frac{x^{frac{1}{2}}}{sqrt{a^3-x^3}} dx | vedclass

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

Question

(C) Let $I = \int \frac{x^{\frac{1}{2}}}{\sqrt{a^3-x^3}} dx$. Substitute $t = x^{\frac{3}{2}}$,then $dt = \frac{3}{2} x^{\frac{1}{2}} dx$,which implie

Vedclass · Accepted Answer

$\frac{2}{3} \sin^{-1}\left(\left(\frac{x}{a}\right)^{\frac{3}{2}}\right)$

Vedclass · Answer

(C) Let $I = \int \frac{x^{\frac{1}{2}}}{\sqrt{a^3-x^3}} dx$. Substitute $t = x^{\frac{3}{2}}$,then $dt = \frac{3}{2} x^{\frac{1}{2}} dx$,which implies $x^{\frac{1}{2}} dx = \frac{2}{3} dt$. Also,$t^2 = (x^{\frac{3}{2}})^2 = x^3$. Substituting these into the integral: $I = \int \frac{\frac{2}{3} dt}{\sqrt{a^3 - t^2}} = \frac{2}{3} \int \frac{dt}{\sqrt{(\sqrt{a^3})^2 - t^2}}$. Using the standard integral formula $\int \frac{du}{\sqrt{A^2 - u^2}} = \sin^{-1}(\frac{u}{A}) + c$: $I = \frac{2}{3} \sin^{-1}\left(\frac{t}{\sqrt{a^3}}\right) + c$. Substituting $t = x^{\frac{3}{2}}$ back: $I = \frac{2}{3} \sin^{-1}\left(\frac{x^{\frac{3}{2}}}{a^{\frac{3}{2}}}\right) + c = \frac{2}{3} \sin^{-1}\left(\left(\frac{x}{a}\right)^{\frac{3}{2}}\right) + c$. Thus,$P(x) = \frac{2}{3} \sin^{-1}\left(\left(\frac{x}{a}\right)^{\frac{3}{2}}\right)$.

If $\int \frac{x^{\frac{1}{2}}}{\sqrt{a^3-x^3}} dx = P(x) + c$,then $P(x) =$

Similar Questions

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 $A+B=$

$\int \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} \, dx$ is equal to

The value of $\int \frac{e^x}{e^x + 1} \,dx$ is

$\int \sqrt{4 \cos ^2 x - 5 \sin ^2 x} \cos x \, dx =$

$ \int \frac{1}{\sqrt{x}+x \sqrt{x}} d x $

Vedclass Test Series

Exam Paper Generator

Online Exam Module