My Math Forum integrals sin and cos

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 July 3rd, 2016, 07:29 AM #1 Member   Joined: Feb 2014 Posts: 91 Thanks: 1 integrals sin and cos I seem to have a hole in my logic. I pulled out a constant from this which was -1/2. I put this constant on the outside of the integral sign. Their results are similar to mine except I have to multiply everything by a constant of -1/2 at the end. I think I messed up when I did U substitute but I haven't seen the mistake yet. Please help
 July 3rd, 2016, 08:45 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,857 Thanks: 2230 Math Focus: Mainly analysis and algebra We can't see your mistake because we can't see your working.
 July 3rd, 2016, 08:52 AM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 2,579 Thanks: 1274 where did you get the constant, $-\dfrac{1}{2}$, from the integral shown ? $\displaystyle \int \sin^2{x}\cos^2{x} \sin{x} \, dx$
 July 3rd, 2016, 10:01 AM #4 Member   Joined: Feb 2014 Posts: 91 Thanks: 1 I've consulted with others. This problem is $\displaystyle \int \sin^2{x}\cos^2{x} \, dx$ When I did my U substitution I let U be equal to cos^2x which was wrong. U should have just been cos(x). I revisited this question multiple time yesterday so I though I needed to ask here. Sorry for wasting your time my apologies.
July 8th, 2016, 04:35 AM   #5
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 Originally Posted by The_Ys_Guy I've consulted with others. This problem is $\displaystyle \int \sin^2{x}\cos^2{x} \, dx$ When I did my U substitution I let U be equal to cos^2x which was wrong. U should have just been cos(x). I revisited this question multiple time yesterday so I though I needed to ask here. Sorry for wasting your time my apologies.
That is not the problem you give in your first post. In your first post you have $\displaystyle \int \sin^2(x)\cos^2(x)sin(x)dx$ which is the same as $\displaystyle \int\sin^3 cos^2(x)dx$.

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