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 March 15th, 2013, 11:39 PM #1 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 4 integration problems Integrate the following $1.\ cos^2x/(sin x)^{1/2} 2.\ (sin x.cos x)^{1/2} 3.\ x/(3-x^4)^{1/2} 4.\ 1/5+4cos x$ I tried to solve these many times, but failed. Please help!
 March 16th, 2013, 01:27 AM #2 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: 4 integration problems 2. integral (sin(x)) ^(1/2) dx is not an elementary integral, it can be reduced to an elliptic integral, so you can not express it with elementary functions. See elliptic integrals. 1. you can reduce it sin^(3/2). it's suspicious that the same situation as in 2. 3. substitute y = x^2 4. it's hard to believe you can't solve it.
 March 16th, 2013, 10:50 AM #3 Newbie   Joined: Feb 2013 Posts: 10 Thanks: 0 Re: 4 integration problems Number four is nothing more than using differentiation rules. The quotient rule is the one you want to use for that one.
 March 16th, 2013, 04:06 PM #4 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 Re: 4 integration problems Is the fourth intended to be 1/(5 + 4cos x)? That would be much more interesting.
March 17th, 2013, 04:52 AM   #5
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Re: 4 integration problems

Quote:
 Originally Posted by skipjack Is the fourth intended to be 1/(5 + 4cos x)? That would be much more interesting.
Yes it's 1/(5+4cos x). I forgot the brackets. I'm terribly sorry.

 March 17th, 2013, 05:06 AM #6 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 Re: 4 integration problems Thanks a lot csak. Your solutions really worked. I'm grateful to you. Please help me with the 4th one. Actually I forgot the bracket which made it look easy to solve.
 March 18th, 2013, 01:19 AM #7 Senior Member   Joined: Oct 2011 From: India ???? Posts: 224 Thanks: 0 Re: 4 integration problems $I= \int \frac{1}{5+4\cos x}dx$ Substitute $t= \tan \left(\frac{x}{2}\right)$, $\cos x= \frac{1-t^2}{1+t^2}$ and $dx= \frac{2}{1+t^2}dt$: $I= \int \frac{1}{5+4 \left( \frac{1-t^2}{1+t^2}\right)} \left(\frac{2}{1+t^2} \right)dt = 2\int \frac{1}{5+5t^2+4-4t^2}dt = 2\int \frac{1}{9+t^2}dt$ $=\frac{2}{3}\arctan \left( \frac{t}{3}\right)+C =\fbox{\displaystyle \frac{2}{3} \arctan \left(\frac{1}{3}\tan \left(\frac{x}{2}\right) \right) +C}$
 March 20th, 2013, 02:57 AM #8 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 Re: 4 integration problems [reply="Etyucan"] Thank you so very much!

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