1. trig integration

integrate 1/sinx(cosx)^4 dx

2. Originally Posted by twilightstr
integrate 1/sinx(cosx)^4 dx
Substitute $u = \cos x \Rightarrow dx = - \frac{du}{\sin x}$. Note that $\sin^2 x = 1 - \cos^2 x = 1 - u^2 \, ....$

3. i still dont know how to solve the problem.

4. Originally Posted by twilightstr
i still dont know how to solve the problem.
Please show how you've tried to use the suggestions I've made. Do you know how to substitute? Where do you get stuck?

5. yes i dont know how to substitute.

6. Originally Posted by twilightstr
yes i dont know how to substitute.
Couldn't you get at least the first line: $\int \frac{1}{\sin x} \, \frac{1}{u^4} \, \left( \frac{-du}{\sin x} \right)$.

Then:

$= - \int \frac{1}{\sin^2 x} \, \frac{1}{u^4} \, du$

$= - \int \frac{1}{1 - u^2} \, \frac{1}{u^4} \, du$

$= - \int \frac{1}{(1 - u^2) u^4} \, du$

Now do a partial fraction decomposition.

If you're doing this sort of question it's expected that you've seen the various techniques.

7. Following up on mr fantastic's last integral we have,

$\frac{1}{\left( 1-{{u}^{2}} \right){{u}^{4}}}=\frac{{{u}^{4}}+\left( 1+{{u}^{2}} \right)\left( 1-{{u}^{2}} \right)}{\left( 1-{{u}^{2}} \right){{u}^{4}}}=\frac{1}{1-{{u}^{2}}}+\frac{1+{{u}^{2}}}{{{u}^{4}}}.$

Besides $\frac{1}{1-{{u}^{2}}}=\frac{(1-u)+(1+u)}{2(1+u)(1-u)}=\frac{1}{2}\left( \frac{1}{1+u}+\frac{1}{1-u} \right)$ and $\frac{1+{{u}^{2}}}{{{u}^{4}}}={{u}^{-4}}+{{u}^{-2}}$ so you got easy integrals to solve!