Trig Substitution again

• Sep 29th 2009, 10:58 AM
Hellacious D
Trig Substitution again
The integral from 0 to a of (x^2)(((a^2)-(x^2))^.5) dx
It must be done using trig substitution. I have it down to
(((a^4)(2sqrt(2))-(a^4))/3 + ((a^4)-((a^4)(4sqrt(2))))/5

Is there any way to simply further?
• Sep 29th 2009, 12:01 PM
Amer
Quote:

Originally Posted by Hellacious D
The integral from 0 to a of (x^2)(((a^2)-(x^2))^.5) dx
It must be done using trig substitution. I have it down to
(((a^4)(2sqrt(2))-(a^4))/3 + ((a^4)-((a^4)(4sqrt(2))))/5

Is there any way to simply further?

$\int_{0}^{a} x^2\sqrt{a^2-x^2} dx$

let $a\sin u = x \Rightarrow a\cos u\cdot du = dx$

$\cos u = \frac{\sqrt{a^2-x^2} }{a}$

$\int a^2\sin ^2u (a\cos u )(a\cos u) du$

$\int a^4 \sin ^2u \cos ^2u\cdot du$

$\int \frac{a^4 \sin ^2(2u)}{4} du$

$\int \frac{a^4 (1-\cos 4u)}{8} du$

$\frac{a^4 (u)}{8} - \frac{a^4\sin 4u}{32}$

but $u=\sin ^{-1} \frac{x}{a}$

$\frac{a^4 \sin ^{-1} \frac{x}{a}}{8} - \frac{a^4\sin 4(\sin ^{-1} \frac{x}{a})}{32} \mid ^{a}_{0}$

$\frac{a^4 \sin ^{-1} (1)}{8} - \frac{a^4 \sin 4(\sin ^{-1} (1))}{32} - \frac{a^4 \sin ^{-1} (0)}{8} + \frac{a^4 \sin 4(\sin ^{-1} (0))}{32}$

$\sin ^{-1} (1) = \frac{\pi}{2}$

$\sin ^{-1} (0) = 0$