# Good examples on Integrals

• August 9th 2007, 08:59 AM
wizzler
Good examples on Integrals
Hey!

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

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

with partial integration/substitution methods?
• August 9th 2007, 09:11 AM
Krizalid
First rewrite the integrand as follows

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

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

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

Just post if you cannot take it from there.
• August 9th 2007, 09:14 AM
ThePerfectHacker
Just be careful this is a "Improper Integral of the 2nd Type".
• August 9th 2007, 09:34 AM
wizzler
Quote:

Originally Posted by Krizalid
First rewrite the integrand as follows

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

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

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

Just post if you cannot take it from there.

so $\int_1^2\frac1{u(u-1)^2}\,du$ can be put as

$u \int_1^2\frac1{(u-1)^2}\,du$ and $\frac1{(u-1)^2}\$ is $ln((u-1)^2)$ or am i thinking it totally wrong now?
• August 9th 2007, 09:40 AM
wizzler
neuu of course...

i part integrate it so i get $\frac{a}{u^2} + \frac{b}{u} + \frac{c}{u + 1}$
• August 9th 2007, 10:05 AM
Krizalid
Quote:

Originally Posted by wizzler
so $\int_1^2\frac1{u(u-1)^2}\,du$ can be put as

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

This step is not correct 'cause "u" it's a function of "x", not a constant.

We have


\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.