Originally Posted by

**twostep08** ive tried everything i could think of and cant get a proper u.

x^2 (sq.rt.(x+1)) dx

By parts! $\displaystyle u=x^2\,,\,u'=2x\,,\,\,v'=\sqrt{x+1}\,,\,v=\frac{2} {3}(x+1)^{3\slash2}\,,\,\,\,so:$

$\displaystyle \int x^2\sqrt{x+1}\,dx=\frac{2}{3}x^2(x+1)^{3\slash2}-\frac{4}{3}\int x\sqrt{x+1}\,dx$....now you continue (again by parts)!

(sin x) dx/(sq.rt.(cos^3 x))

$\displaystyle \int\frac{\sin x}{\sqrt{\cos^3x}}dx=\int\sin x\left(\cos^{-3\slash 2}\!(x)\right)dx$ , and this is an integral of the form $\displaystyle \int f'(x)f(x)^n\,dx=\frac{f(x)^{n+1}}{n+1} + C$

OR you can make the substitution $\displaystyle u=\cos x\Longrightarrow du=-\sin x \,dx$ and etc.

sin (sq.rt.(x+1)) dx/ (sq.rt. (x+1) from 3 to 8

$\displaystyle u=\sqrt{x+1}\Longrightarrow du=\frac{dx}{2\sqrt{x+1}}\,,\,\,and\,\,x=3\Longrig htarrow u=\sin 2$ and etc.

Tonio

they dont seem like they should cause too much problems, but for the life of me, i cant figure them out

any help is appreciated, as always