# Math Help - Primitive of Reciprocal of sin^3 ax

1. ## Primitive of Reciprocal of sin^3 ax

While I do some investigation of the complicated induction work on no. 335 ...

No. 352 in my list (there were a load of easy sine formulas):

$\int \frac {\mathrm d x} {\sin^3 a x}$

which is supposed to work out to:

$-\frac {\cos a x} {2 a \sin^2 a x} + \frac 1 {2 a} \ln \left\vert{\tan \frac {a x} 2}\right\vert$

but is going a bit astray.

I tried Parts, with $u = \frac 1 {\sin a x}$ and got so far:

$-\frac {\cos a x} {a \sin^2 a x} - \int \frac {\cos^2 a x} {\sin^3 a x} \ \mathrm d x$

Hints?

2. ## Re: Primitive of Reciprocal of sin^3 ax

your first answer is correct

3. ## Re: Primitive of Reciprocal of sin^3 ax

oh you're having trouble getting there?

4. ## Re: Primitive of Reciprocal of sin^3 ax

The last part is actually not as bad:
S cos^2(ax)/sin^3(ax) dx right? Rewrite as
S cot^2(ax)*csc(ax) dx
Letting u=csc(ax), we have du=-a*cot^2(ax)dx. Then, cot^2(ax)dx = (-1/a)du Now the integral becomes
S u*(-1/a)du = (-1/a) S u du = (-1/a) u^2/2 = (-1/a) csc^2(ax)/2 = (-1/2a) csc^2(ax) = -1/[2a*sin^2(ax)].

Hmm... the natural logarithm doesn't show up anywhere. Maybe you should have started with S csc^3(ax) dx instead...

5. ## Re: Primitive of Reciprocal of sin^3 ax

Originally Posted by dennydengler
The last part is actually not as bad:
S cos^2(ax)/sin^3(ax) dx right? Rewrite as
S cot^2(ax)*csc(ax) dx
Letting u=csc(ax), we have du=-a*cot^2(ax)dx. Then, cot^2(ax)dx = (-1/a)du Now the integral becomes
S u*(-1/a)du = (-1/a) S u du = (-1/a) u^2/2 = (-1/a) csc^2(ax)/2 = (-1/2a) csc^2(ax) = -1/[2a*sin^2(ax)].

Hmm... the natural logarithm doesn't show up anywhere. Maybe you should have started with S csc^3(ax) dx instead...
Yep, I'll try that properly -- except that $\frac d{dx} \csc ax \ne a \cot^2 ax$ ... this appears to be where the ln comes in.

6. ## Re: Primitive of Reciprocal of sin^3 ax

Oh oops, bad mistake.

8. ## Re: Primitive of Reciprocal of sin^3 ax

Originally Posted by Matt Westwood
Yep, I'll try that properly -- except that $\frac d{dx} \csc ax \ne a \cot^2 ax$ ... this appears to be where the ln comes in.
I think the substitution to use is $u = \sin^2 ax$ and then it turns into a simple integral $\int \frac {1 - u^2}u \ \du$ unless I've blundered.

9. ## Re: Primitive of Reciprocal of sin^3 ax

Originally Posted by tinspire
I'm wishing I didn't know about this now, it looks like it might take all the fun away ...

10. ## Re: Primitive of Reciprocal of sin^3 ax

sorry, just wanted to help. i though maybe you would be able to see where you went wrong in your steps.

11. ## Re: Primitive of Reciprocal of sin^3 ax

Originally Posted by tinspire
sorry, just wanted to help. i though maybe you would be able to see where you went wrong in your steps.
Yer alrite pal ... British sense of humour.

12. ## Re: Primitive of Reciprocal of sin^3 ax

As mentioned earlier your new integral is

$- \int \csc ax \cot^2 ax \, dx = - \int \csc ax (\csc^2 ax -1) \, dx$
$= - \int \csc^3 ax \, dx + \int \csc ax \, dx$

Now move over the $csc^3 ax$ integral to the other side of the equation and evaluate the single $csc ax$ integral.

13. ## Re: Primitive of Reciprocal of sin^3 ax

Originally Posted by Matt Westwood
Yer alrite pal ... British sense of humour.
oh ok. nevermind, not sorry