I'm going to rewrite the function with parentheses to what I assume it's supposed to look like:
∫(x*sin(x))/(1+cos(2x)) dx, 0, pi/2)
It's helpful to know trig identities.
cos(2x) = cos(x+x) = cos^2(x) - sin^2(x)
Take a look at just the denominator:
1 + cos^2(x) - sin^2(x) = (1 - sin^2(x)) + cos^2(x) = cos^2(x) + cos^2(x) = 2*cos^2(x)
So now you have:
∫x*sin(x))/(2*cos^2(x)) dx = 1/2∫x*sin(x))/(cos^2(x)) dx
Let's rewrite this as:
1/cos(x) is sec(x)
sin(x)/cos(x) = tan(x)
So now we've got:
This would be easy to evaluate if that pesky little x term wasn't there... fortunately, though, we can integrate by parts.
u = x, and dv = sec(x)tan(x)
du = 1*dx, v = sec(x)
And plug in our values for:
u*v - ∫v*du
We arrive at:
x*sec(x) - ∫sec(x) dx
x*sec(x) - ln(tan(x) + sec(x))
Don't forget the 1/2 term:
1/2*[x*sec(x) - ln(tan(x) + sec(x))]
You'll have to check your upper limit because pi/2, when plugged into both sec(x) and tan(x), results in 1/0 which is undefined. You'll also be able to check that from your original equation, since (1 + cos(2*pi/2)) = (1 + cos(pi)) = 1 - 1 = 0. Since that 0 is in the denominator, the function is undefined.
This integral is improper since it approaches infinity at its upper limit.