# Good examples on Integrals

Printable View

• Aug 9th 2007, 07:59 AM
wizzler
Good 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 AM
Krizalid
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 AM
ThePerfectHacker
Just be careful this is a "Improper Integral of the 2nd Type".
• Aug 9th 2007, 08:34 AM
wizzler
Quote:

Originally Posted by Krizalid
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.

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

$\displaystyle u \int_1^2\frac1{(u-1)^2}\,du$ and $\displaystyle \frac1{(u-1)^2}\$ is $\displaystyle ln((u-1)^2)$ or am i thinking it totally wrong now?
• Aug 9th 2007, 08:40 AM
wizzler
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 AM
Krizalid
Quote:

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

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

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

We have

\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.