# how can i integrate these using substitution

• Nov 12th 2009, 08:16 AM
twostep08
how can i integrate these using substitution
ive tried everything i could think of and cant get a proper u.

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

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

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

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
• Nov 12th 2009, 08:37 AM
tonio
Quote:

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

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

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

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

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

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

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

.
• Nov 12th 2009, 08:53 AM
Plato
Quote:

Originally Posted by twostep08
ive tried everything i could think of and cant get a proper u.
x^2 (sq.rt.(x+1)) dx

$\begin{gathered}
u = x + 1\; \Rightarrow \;x = u - 1 \hfill \\
x^2 \sqrt {x + 1} \; \Rightarrow \;\left( {u - 1} \right)^2 \sqrt u = u^{\frac{5}
{2}} - 2u^{\frac{3}
{2}} + u^{\frac{1}
{2}} \hfill \\
\end{gathered}$
• Nov 12th 2009, 10:32 AM
twostep08
thanks guys...i knew how to do the first by parts, but we had to do it by u-sub...either way, i understand all 3 now--thanks!