# Integration of 1/sin^4(x)

• Sep 27th 2012, 07:23 PM
mathimaticool
Integration of 1/sin^4(x)
I need help integrating this function. I've done integration by parts many times, and it seems to be an infinite loop. Please help, would be greatly appreciated! :)
• Sep 27th 2012, 07:31 PM
TheEmptySet
Re: Integration of 1/sin^4(x)
Quote:

Originally Posted by mathimaticool
I need help integrating this function. I've done integration by parts many times, and it seems to be an infinite loop. Please help, would be greatly appreciated! :)

$\frac{1}{\sin^4(x)}=\csc^4(x)=\csc^2(x)\csc^2(x)=( 1+\cot^2(x))\csc^2(x)$

Now let $u=\cot(x) \implies du=-\csc^2(x)dx$

Can you finish from here?
• Sep 28th 2012, 10:13 PM
mathimaticool
Re: Integration of 1/sin^4(x)
Ahh ok. But when I get to

-cot^2(x) - integral(csc^2(x)cot(x) dx)

I don't know what to do. Thanks a lot
• Sep 29th 2012, 04:15 AM
TheEmptySet
Re: Integration of 1/sin^4(x)
Quote:

Originally Posted by mathimaticool
Ahh ok. But when I get to

-cot^2(x) - integral(csc^2(x)cot(x) dx)

I don't know what to do. Thanks a lot

So you have

$\int (1+\cot^2(x))\csc^2(x) dx$

If you let $u=\cot(x) \implies du = -\csc^2(x)$

This gives

$\int (1+\cot^2(x))\csc^2(x) dx=\int (1+u^2)(-du)=-\int (1+u^2)du$
• Sep 29th 2012, 04:31 PM
mathimaticool
Re: Integration of 1/sin^4(x)
AHHH thanks soo much! Got it!