Integrating $\frac{x}{1 + \sin(x) \cos(x)}$ from 0 to $\pi$

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<br /> \int_0^{\pi} \frac{x}{1+\sin(x)\cos(x)} dx<br />
 
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how does latex in this forum works? how to get rid of the 18416 ?

testing

\frac{1}{2}

\int_0^{\pi} \frac{x}{1+\sin(x)\cos(x)} dx
 
Why is this a hard integral? Post your work please
 
what is the method used?
 
This is not a competition. I had the answer the minute I looked at the problem to make sure it existed. The point is to teach him how to find the solution by looking at his attempt at the solution. Please don't randomly post an answer which does not help him in any way.


To OP: for what its worth, sin(2*x) = 2*sin(x)*cos(x). You can try u=2*x, du = 2 dx
 
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cronxeh said:
This is not a competition. I had the answer the minute I looked at the problem to make sure it existed. The point is to teach him how to find the solution by looking at his attempt at the solution. Please don't randomly post an answer which does not help him in any way.

Exactly. I've deleted the posts with answers in them. This is most likely homework, so giving the OP the answer is cheating. I'm moving this thread to Homework Help, Calculus and Beyond.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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