int_0^a {x^4 sqrt{a^2 - x^2}} ,dx = | vedclass

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

Question

(B) Let $I = \int_0^a x^4 \sqrt{a^2 - x^2} \,dx$. Put $x = a \sin 	heta$,then $dx = a \cos 	heta \,d	heta$. When $x = 0$,$	heta = 0$. When $x = a$

Vedclass · Accepted Answer

$\frac{\pi}{32} a^6$

Vedclass · Answer

(B) Let $I = \int_0^a x^4 \sqrt{a^2 - x^2} \,dx$. Put $x = a \sin 	heta$,then $dx = a \cos 	heta \,d	heta$. When $x = 0$,$	heta = 0$. When $x = a$,$	heta = \frac{\pi}{2}$. Substituting these into the integral: $I = \int_0^{\pi/2} (a \sin 	heta)^4 \sqrt{a^2 - a^2 \sin^2 	heta} \cdot (a \cos 	heta) \,d	heta$ $I = \int_0^{\pi/2} a^4 \sin^4 	heta \cdot (a \cos 	heta) \cdot (a \cos 	heta) \,d	heta$ $I = a^6 \int_0^{\pi/2} \sin^4 	heta \cos^2 	heta \,d	heta$. Using the reduction formula $\int_0^{\pi/2} \sin^m 	heta \cos^n 	heta \,d	heta = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$: $I = a^6 \cdot \frac{\Gamma(\frac{5}{2}) \Gamma(\frac{3}{2})}{2 \Gamma(\frac{4+2+2}{2})} = a^6 \cdot \frac{(\frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi}) \cdot (\frac{1}{2} \cdot \sqrt{\pi})}{2 \cdot \Gamma(4)}$ $I = a^6 \cdot \frac{\frac{3}{8} \pi}{2 \cdot 6} = a^6 \cdot \frac{3\pi}{96} = \frac{\pi}{32} a^6$.

$\int_0^a {x^4 \sqrt{a^2 - x^2}} \,dx = $

Similar Questions

If $g(1) = g(2)$,then the value of $\int_{1}^{2} [f\{g(x)\}]^{-1} f'\{g(x)\} g'(x) dx$ is:

The value of the integral $\int \limits_{0}^{1} \frac{\sqrt{x} \, dx}{(1+x)(1+3 x)(3+x)}$ is:

The value of $\int_0^{\infty} \frac{dx}{(x^2+4)(x^2+9)}$ is

$\int_{-2}^3 |1-x^2| dx =$

Evaluate the definite integral $\int_{2}^{3} \frac{d x}{x^{2}-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module