#\int_(0)^(15)x^2\sqrt(a^2-x^2)dx#?

I got as far as #\inta^4\sin^4(\theta)\cos^2(theta)d\theta# from trigonometric substitution with

#x=a\sin(\theta)#
#dx=a\cos(\theta)d\theta#

NOTE

The original integral looked like this

#\int_(0)^(a)x^2\sqrt(a^2-x^2)dx#

In this version, the upper integration limit is #a#, not #15#.

1 Answer
Apr 15, 2018

#int_0^a x^2sqrt(a^2-x^2)dx = (a^4pi)/16#

Explanation:

Evaluate:

#int_0^a x^2sqrt(a^2-x^2)dx #

Substitute:

#x= asint#

#dx = a costdt#

with #t in [0,pi/2]#

so that:

#int_0^a x^2sqrt(a^2-x^2)dx = int_0^(pi/2) a^2 sin^2t sqrt(a^2-a^2 sin^2t)acostdt#

#int_0^a x^2sqrt(a^2-x^2)dx = a^4 int_0^(pi/2) sin^2t sqrt(1- sin^2t)costdt#

For #t in [0,pi/2]# the cosine os positive, so:

#sqrt(1- sin^2t) = cost#

and then:

#int_0^a x^2sqrt(a^2-x^2)dx = a^4 int_0^(pi/2) sin^2t cos^2tdt#

Use now the trigonometric identities:

#sin 2theta = 2 sin theta cos theta#

#2sin^2 theta = 1- cos theta#

so:

#sin^2t cos^2tdt = 1/4 sin^2 2t = 1/8(1-cos4t)#

and:

#int_0^a x^2sqrt(a^2-x^2)dx = a^4/8 int_0^(pi/2) (1-cos4t)dt#

Using now the linearity of the integral:

#int_0^a x^2sqrt(a^2-x^2)dx = a^4/8 (int_0^(pi/2) dt - int_0^(pi/2) cos4tdt)#

#int_0^a x^2sqrt(a^2-x^2)dx = a^4/8 [t - (sin 4t)/4]_0^(pi/2)#

#int_0^a x^2sqrt(a^2-x^2)dx = (a^4pi)/16#