Question

integrate (x+sinx)/(1+cosx) from 0 to pi/2

Answer 1

hmm im not seeing an easy way to do this.

i would split it into 2 fractions and integrate separately
x/(1+cos) ---> use integration by parts

sin/(1+cos) --> use substitution

Answer 2

Looks like it might involve this substitution: tan(x/2) = t

So right away, you have x = 2arctan(t), as well as
1/2 sec²(x/2) dx = dt
dx = 2 cos²(x/2) dt

Making use of some trig identities, you'll find a few expressions you'll need:
cosx = cos(2 (x/2)) = cos²(x/2) - sin²(x/2)
sinx = sin(2 (x/2)) = 2 sin(x/2) cos(x/2)

Use the initial sub to draw yourself a triangle that satisfies the equation tan(x/2) = t. You'll find yourself with a triangle that has
sin(x/2) = t/(t² + 1)
cos(x/2) = 1/(t² + 1)

So, you get
cosx = (1 - t²)/(t² + 1)²
sinx = 2t/(t² + 1)²

Returning to the differential substitution, you have
dx = 2 [(1 - t²)/(t² + 1)²]² dt

So the integral becomes
Int[ (x + sinx)/(1 + cosx) dx ] = ...
= Int[ (2arctan(t) + 2t/(t² + 1)²) / (1 + (1 - t²)/(t² + 1)²) * 2 [(1 - t²)/(t² + 1)²]² dt ]

Hard to make out? Here's what it should look like: