# Trigonometric Substitution

I am having some trouble with a homework problem that reads:"Use the subsituition $x=\sin(t)$ to evaluate the $\int\sqrt{36-x^2}dx$. I would really appreciate if someone could help me with this problem!

The substitution should be $x=6\sin(t)$. If we substitute $6\sin(t)$ for $x$ in the expression, you get $$\sqrt{36-x^2}=\sqrt{36-(6\sin t)^2}=\sqrt{36-36\sin^2t}=\sqrt{36(1-\sin^2t)}=6\sqrt{\cos^2t}=6\cos t$$ That's all there is to it. Do the substitution, then apply the appropriate pythagorean identity.