1. ## Trig Integration

I have to integrate sin^2xcos^5x dx

I've come to the answer of -t^6/6 +t^11/11 + C where t= cosx

Im worried its incorrect

2. Originally Posted by qwerty10
I have to integrate sin^2xcos^5x dx

I've come to the answer of -t^6/6 +t^11/11 + C where t= cosx

Im worried its incorrect
If you ever need to check your integration, use Wolfram.

3. Using the identity $\displaystyle \cos^2(x) = 1-\sin^2(x)$ and writing $\displaystyle \cos^4(x) = (\cos^2(x))^2$ the question can be rewritten as $\displaystyle \sin^2(x)(1-\cos^2(x))^2\cos(x)$
Let $\displaystyle u = \sin(x)$ so that $\displaystyle du = \cos(x)\, dx \implies dx = \dfrac{du}{\cos(u)}$
$\displaystyle \int u^2(1-u^2)^2 du$ and expanding the brackets we get $\displaystyle \int u^2(1-2u^2+u^4)\ du = \int u^2-2u^4+u^6\ du$