x1-cosxsinx-xcosxdx

We can start by simplifying the expression in the numerator:

x(1-cosx) = x - xcosx

Now we can rewrite the denominator using the identity sinx-xcosx=-xcosx/sinx:

sinx-xcosx = sinx - (sinx-xcosx) = 2xcosx - sinx

Substituting these expressions into the original integral, we get:

∫(x-xcosx)/(2xcosx-sinx) dx

Now we can use a substitution u = 2xcosx - sinx, which gives us du/dx = -2xsinx - cosx. Rearranging this expression, we get:

xsinx = (2xcosx - u)/2

Substituting these expressions into the integral, we get:

∫[(2xcosx-u)/2-u/2] / u du/dx dx

Simplifying the numerator, we get:

∫(3u/4 - 2xcosx) / u (-2xsinx - cosx) dx

Now we can cancel out the x terms and simplify the expression further:

∫(3/4 - u/2cosx) / u (-2sinx - 1/2cosx) dx

Using partial fractions, we can write:

(3/4 - u/2cosx) / u (-2sinx - 1/2cosx) = A/u + B/(-2sinx - 1/2cosx)

Multiplying both sides by the denominator, we get:

3/4 - u/2cosx = A(-2sinx - 1/2cosx) + B(u)

Substituting x = π/2, we get:

3/4 - 0 = B(2)

B = 3/8

Substituting x = 0, we get:

3/4 - u/2 = A(-1/2)

A = -3/4u

Substituting these values back into the partial fractions expression, we get:

(-3/4u/u) + (3/8)/(2sinx + 1/2cosx)

Simplifying, we get:

-3/4∫(1/u) du + 3/4∫(1/(2sinx + 1/2cosx)) dx

The first integral is ln|u|, and the second integral can be evaluated using a substitution v = 2sinx + 1/2cosx, which gives us:

∫(1/v) dv = ln|v| = ln|2sinx + 1/2cosx|

Substituting u = 2xcosx - sinx and simplifying, we get:

∫(x - ln|2xcosx - sinx|)/2 dx

This is the final answer.