Hey!

Can anyone give me good examples on how to solve integrals like:

$\displaystyle \int_{0}^{\pi/2} \frac{cos x}{sin^2x + sin^3 x} dx$

with partial integration/substitution methods?

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- Aug 9th 2007, 07:59 AMwizzlerGood examples on Integrals
Hey!

Can anyone give me good examples on how to solve integrals like:

$\displaystyle \int_{0}^{\pi/2} \frac{cos x}{sin^2x + sin^3 x} dx$

with partial integration/substitution methods? - Aug 9th 2007, 08:11 AMKrizalid
First rewrite the integrand as follows

$\displaystyle \int_0^{\pi/2}\frac{\cos x}{\sin^2x+\sin^3x}\,dx=\int_0^{\pi/2}\frac{\cos x}{\sin^2x(1+\sin x)}\,dx$

Then set $\displaystyle u=1+\sin x\implies du=\cos x\,dx$, which yields

$\displaystyle \int_1^2\frac1{u(u-1)^2}\,du$

Just post if you cannot take it from there. - Aug 9th 2007, 08:14 AMThePerfectHacker
Just be careful this is a "Improper Integral of the 2nd Type".

- Aug 9th 2007, 08:34 AMwizzler
- Aug 9th 2007, 08:40 AMwizzler
neuu of course...

i part integrate it so i get $\displaystyle \frac{a}{u^2} + \frac{b}{u} + \frac{c}{u + 1}$ - Aug 9th 2007, 09:05 AMKrizalid
This step is not correct 'cause "u" it's a function of "x", not a constant.

We have

$\displaystyle

\begin{aligned}

\frac1{u(u-1)^2}&=\frac{u^2-2u+1+u-u^2+u}{u(u-1)^2}\\&=\frac{(u-1)^2+u-u(u-1)}{u(u-1)^2}\\&=\frac1u+\frac1{(u-1)^2}-\frac1{u-1}

\end{aligned}

$

Kill it now.