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.
Your integral.
You can even ask it to show you the intermediate steps
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 $\displaystyle du = \cos(x)\, dx \implies dx = \dfrac{du}{\cos(u)}$
Now putting our subs into the integral we get
$\displaystyle \int u^2(1-u^2)^2 du$ and expanding the brackets we get $\displaystyle \int u^2(1-2u^2+u^4)\ du = \int u^2-2u^4+u^6\ du$