# Thread: Integration - powers of trig functions

1. ## Integration - powers of trig functions

$\displaystyle \int 33\frac{cos^5(x)}{\sqrt{sin(x)}}$

Should I begin by splitting up the numerator? Multiplying by the conjugate? I've tried those two and some others to no avail.

THanks

2. Originally Posted by silencecloak
$\displaystyle \int 33\frac{cos^5(x)}{\sqrt{sin(x)}}$

Should I begin by splitting up the numerator? Multiplying by the conjugate? I've tried those two and some others to no avail.

THanks
Put
$\displaystyle sin(x) = t$

$\displaystyle cos(x)dx = dt$

$\displaystyle 33\int \frac{cos^5(x)dx}{\sqrt{sin(x)}}$

$\displaystyle = 33\int \frac{cos^4(x)dt}{\sqrt{t}}$

Now
$\displaystyle cos^4(x)$

$\displaystyle = (1 -sin^2(x))^2$

$\displaystyle = (1 - t^2)^2$

$\displaystyle = 1 + t^4 - 2t^2$

Hence integration becomes

$\displaystyle = 33\int \frac{(1 + t^4 - 2t^2 )dt}{\sqrt{t}}$

3. Interesting...makes sense...I wish I could see this kind of stuff as I'm working the problems

Edit: now thats not becoming a trig substitution is it?

4. Originally Posted by silencecloak
Interesting...makes sense...I wish I could see this kind of stuff as I'm working the problems

Edit: now thats not becoming a trig substitution is it?
First of all
Congrats on Becoming a Senior Member and on your 100 posts

Yes its trignometric substitution

First of all
Congrats on Becoming a Senior Member and on your 100 posts

Yes its trignometric substitution

Thanks haha

Is there another way to do this problem without trig sub? The chapter this homework is for has not covered that topic yet.

6. Originally Posted by silencecloak
Thanks haha

Is there another way to do this problem without trig sub? The chapter this homework is for has not covered that topic yet.
On a second thought, its something similar to your last question

On a second thought, its something similar to your last question
By last question, do you mean this post?

http://www.mathhelpforum.com/math-he...ig-powers.html

8. Yes If you do it that way , you should to do this one as well
Reason: You did some similar substitution there

Yes If you do it that way , you should to do this one as well
Reason: You did some similar substitution there
Ok, I will go back and look at this problem with that technique in mind, darn algebra

10. $\displaystyle 33\int sin^{-1/2}(x)cos^4(x)cos(x)dx$

$\displaystyle 33\int sin^{-1/2}(x)[1-sin^2(x)]^2cos(x)dx$

$\displaystyle u = sin(x)$
$\displaystyle du = cos(x)$

$\displaystyle 33\int u^{-1/2}(1-u^2)^2du$

Is this the right way?

Edit: It's the right way, thanks for the help