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

Integrate the function $\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$.

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

Consider the following statements:
Assertion $(A):$ $\frac{1}{x^2 + a^2}$ can be integrated by a substitution $x = a \tan \theta$.
Reason $(R):$ Because all integrands are integrated by the method of substitution only.
Which of the following is correct?

Integrate the function: $\frac{1}{x \sqrt{ax - x^{2}}} \quad \left[ \text{Hint: } x = \frac{a}{t} \right]$

If $\int \frac{dx}{5 + 4\cos x} = \lambda \tan^{-1} \left( m \tan \frac{x}{2} \right) + C$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module