Question

`\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`.

Answer 1

`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`