December 2nd, 2012, 03:27 PM  #1 
Member Joined: Apr 2012 Posts: 72 Thanks: 3  solution of integral
cot(cos^1 x)

December 2nd, 2012, 06:36 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,168 Thanks: 472 Math Focus: Calculus/ODEs  Re: solution of integral
Rewrite the integrand using the identity: then use the substitution: to get: 
December 2nd, 2012, 09:38 PM  #3 
Member Joined: Apr 2012 Posts: 72 Thanks: 3  Re: solution of integral
Thanks.

December 2nd, 2012, 10:38 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,168 Thanks: 472 Math Focus: Calculus/ODEs  Re: solution of integral
Use the powerrule for integration: where along with the antiderivative form of the fundamental theorem of calculus: where . 
December 3rd, 2012, 12:33 AM  #5 
Member Joined: Apr 2012 Posts: 72 Thanks: 3  Re: solution of integral
Can someone please post the way from start to the end? 
December 3rd, 2012, 12:40 AM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,168 Thanks: 472 Math Focus: Calculus/ODEs  Re: solution of integral
You cannot evaluate the resulting definite integral using the rules I posted?

December 3rd, 2012, 12:51 AM  #7 
Member Joined: Apr 2012 Posts: 72 Thanks: 3  Re: solution of integral
I tried, but I have an error somewhere. I got a division by zero.

December 3rd, 2012, 01:05 AM  #8 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,168 Thanks: 472 Math Focus: Calculus/ODEs  Re: solution of integral 
December 3rd, 2012, 01:23 AM  #9 
Member Joined: Apr 2012 Posts: 72 Thanks: 3  Re: solution of integral
How did you do this first step? 
December 3rd, 2012, 01:29 AM  #10 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,168 Thanks: 472 Math Focus: Calculus/ODEs  Re: solution of integral is an angle in a right triangle where the ratio of the side adjacent to that angle and the hypotenuse is , so we may say that the adjacent side is and the hypotenuse is 1. By the Pythagorean theorem, we may then say the side opposite the angle is . Since the cotangent function by definition is the ratio of the adjacent to the opposite, we then have: When I was first learning this, I would draw a right triangle so that I could clearly see what I needed to find. 

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