Tricky Trig Integration: How to Tackle Powers and Parts?

[graycolor]

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.

 

[Nick89]

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]

Sorry, its attached now.

 

[graycolor]

So you can't see what I attached right now.

 

[graycolor]

Okay here it is: Integral of (sin(14x))^3/cot(14x) respects to x from 0 to pi/42

 

[Nick89]

Don't forget your trig identities!

A few you might find particularly useful here:

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

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]

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]

EDIT
Oh dang, I misread your equation to be * cot... instead of / cot...

Let me have a look again. Hold on.

 

[snipez90]

I am confused here, is the integrand

$$\frac{sin^{3}(14x)}{cot(14x)}$$

?

 

[Nick89]

I am confused here, is the integrand

$$\frac{sin^{3}(14x)}{cot(14x)}$$

?

Yeah I think it is, but I misread it to be $$\sin^{3}(14x) \cot(14x)$$

 

[graycolor]

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]

There use to be a mark as solved feature can't seem to find it.