integrate with limits 0 to pie
1) x3log(sinx)
2) x/{1+(sinxcosx)}
Use the properties of definite integral.Put the pie-x in place of x and open the cube term.From the cube term you will get four terms and solve each of them separately.The first integral is pie3logsinx.This integral is standard between limits 0 to pie/2 and is equal to pie/2 log(1/2).Other term will be like that of the question .Third integral will be -3(pie)2xlogsinx and here also put pie-x in place of x.You will find that it turns out to be like logsinx and logsin(pie-x)=logsinx .Then you are left with the fourth integral that is x2logsinx between limits 0 to pie.This integral is a problem and if you can find it's solution then the question is solved.
For, \( \int_{0}^{ \pi } \frac{x}{1+sinxcosx} \,dx\) look at the embedded flash below (page 1) or if you do not see it click here and see page 1.