How do I integrate ∫ xcos^2(x) dx ?

First, recall that when integrating, squared trigonometric functions often cause issues. Therefore, use the identity: cos^2(x) = (cos(2x) + 1)/2 to remove this power, giving the integral: ∫ (xcos(2x))/2  + x/2 dx.

This can be split into two integrals: ∫ (xcos(2x))/2 dx + ∫ x/2 dx

The second integral [ ∫ x/2 dx] is simple, integrating to x^2/4

For the first, trickier integral [∫ (xcos(2x))/2 dx], notice that this integral appears to be the multiple of two functions, so integration by parts is likely to come in handy.  Set u = x/2 and dv/dx = cos(2x). Then du/dx = 1/2 and v = ∫ cos(2x) dx = (sin(2x))/2 

Use the integration by parts formula: integral = uv - ∫v(du/dx) dx = xsin(2x)/4 - ∫sin(2x)/4 = xsin(2x)/2 +cos(2x)/8

Put everything together, giving a final integral = xsin(2x)/4 +cos(2x)/8 + x^2/4 + c (don't forget to add the arbitrary constant!)