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Integration of trig powers

by graycolor
Tags: integration, powers, trig
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graycolor
#1
Sep9-08, 12:54 PM
P: 34
I need help I don't even know where to begin. The problem is attached, does this problem deal with integration by parts or should break sec down and some how get cot.
Attached Thumbnails
trig.png  
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Nick89
#2
Sep9-08, 12:56 PM
P: 550
You seem to have forgotten the attachment?

Either way, you might want to consider trying to type it out in latex since attachments can take some time to be approved.
graycolor
#3
Sep9-08, 12:57 PM
P: 34
Sorry, its attached now.

graycolor
#4
Sep9-08, 12:57 PM
P: 34
Integration of trig powers

So you can't see what I attached right now.
graycolor
#5
Sep9-08, 01:05 PM
P: 34
Okay here it is: Integral of (sin(14x))^3/cot(14x) respects to x from 0 to pi/42
Nick89
#6
Sep9-08, 01:59 PM
P: 550
Don't forget your trig identities!

A few you might find particularly useful here:

[tex]\cot x = \frac{\cos x}{\sin x}[/tex]
[tex]\sin^2 x = \frac{1 - \cos 2x}{2}[/tex]
[tex]\cos x \cos y = \frac{1}{2} \left[ \cos(x-y) + \cos(x+y) \right][/tex]
[tex]\cos(-x) = \cos x[/tex]

I think those are all you will need. Use them too rewrite the formula beneath the integral, you will see that it becomes much easier, as is very often the case!
graycolor
#7
Sep9-08, 02:18 PM
P: 34
After rewriting I get sec(14x)^3/3*tan(14x)dx letting my u equal sec(14x) and du=sec(14x)tan(14x) I get the integral of U^2du and the final answer comes to U^3/3... am I correct. Then plugging in sec back in, I get the final answer of sec(14x)^3/3...the funny thing is when I take the derivative I get 14sec(14x)^2*sec(14x)*tan(14x) and not the original equation that should be sec(14x)^2*sec(14x)tan(14x) I must be missing something.
Nick89
#8
Sep9-08, 02:42 PM
P: 550
EDIT
Oh dang, I misread your equation to be * cot... instead of / cot...

Let me have a look again. Hold on.
snipez90
#9
Sep9-08, 02:44 PM
P: 1,104
I am confused here, is the integrand

[tex]\frac{sin^{3}(14x)}{cot(14x)}[/tex]

?
Nick89
#10
Sep9-08, 02:46 PM
P: 550
Quote Quote by snipez90 View Post
I am confused here, is the integrand

[tex]\frac{sin^{3}(14x)}{cot(14x)}[/tex]

?
Yeah I think it is, but I misread it to be [tex]\sin^{3}(14x) \cot(14x)[/tex]
graycolor
#11
Sep9-08, 02:59 PM
P: 34
sorry guys I meant sec(14x)^3/cot(14x)...I figured it out myself for those interested here it is... after rewriting we bring up cot and make it a tan so its now sec(14x)^3*tan(14x) letting u=sec(14x) are du is 14sec(14x)tan(14x) missing things around our equation now looks like 1/14*U^2du after integration its U^3/42 our final answer is sec(14x)^3/42 then just do the fundamental theorem. Sorry for typing the wrong problem.
graycolor
#12
Sep9-08, 03:00 PM
P: 34
There use to be a mark as solved feature can't seem to find it.


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