# Thread: Integration with Trig

1. ## Integration with Trig

Hi,

If you look on the attachment, I cannot understand how to get from the top equation to the second one. I mean, how does 3sin^2(x).cos(x) become just sin^3(x) ???

Is it something to do with double angle formulae?

Somebody please explain!!

Thanks

2. Originally Posted by CSG18
Hi,

If you look on the attachment, I cannot understand how to get from the top equation to the second one. I mean, how does 3sin^2(x).cos(x) become just sin^3(x) ???

Is it something to do with double angle formulae?

Somebody please explain!!

Thanks
Are you familiar with substitution rule in integration? You need to know integration to solve this problem

$\pi \int_0^\frac{\pi}{2} 3sin^{2}x cosx dx$

Let $u = sinx$

then $du = cosx dx$

so, $dx = \frac{du}{cosx}$

Substitute the values of u and dx in your integral, you get:

$\pi \int_0^\frac{\pi}{2} 3u^2 cosx \frac{du}{cosx}$

= $\pi \int_0^\frac{\pi}{2} 3u^2 du$

= $\pi [\frac{3u^3}{3}]_0^\frac{\pi}{2}$

= $\pi [u^3]_0^\frac{\pi}{2}$

Now substitute $u = sin^{2}x$, and you get

$\pi [sin^{3}x]_0^\frac{\pi}{2}$

Then Calculate V with the upper and lower limits provided in the question.