Integration - powers of trig functions

**silencecloak** · February 24th 2009, 12:22 AM

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

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

THanks

**ADARSH** · February 24th 2009, 12:38 AM

Originally Posted by silencecloak

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

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
$<br /> sin(x) = t <br />$

$cos(x)dx = dt$

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

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

Now
$cos^4(x)$

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

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

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

Hence integration becomes

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

Now go ahead

and ask incase of trouble

**silencecloak** · February 24th 2009, 12:41 AM

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?

**ADARSH** · February 24th 2009, 12:45 AM

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

**silencecloak** · February 24th 2009, 12:52 AM

Originally Posted by ADARSH

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.

**ADARSH** · February 24th 2009, 01:18 AM

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

**silencecloak** · February 24th 2009, 09:38 AM

Originally Posted by ADARSH

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

**ADARSH** · February 24th 2009, 09:49 AM

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

**silencecloak** · February 24th 2009, 09:51 AM

Originally Posted by ADARSH

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

**silencecloak** · February 24th 2009, 01:59 PM

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

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

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

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

Is this the right way?

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

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