Integration of 1/sin^4(x)

**mathimaticool** · September 27th 2012, 07:23 PM

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!

**TheEmptySet** · September 27th 2012, 07:31 PM

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?

**mathimaticool** · September 28th 2012, 10:13 PM

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

**TheEmptySet** · September 29th 2012, 04:15 AM

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

$\int (1+\cot^2(x))\csc^2(x) dx=\int (1+u^2)(-du)=-\int (1+u^2)du$

**mathimaticool** · September 29th 2012, 04:31 PM

AHHH thanks soo much! Got it!

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