Evaluate this integral with respect to x xsinxsec^3x?

Recognize that sin(x)sec³(x) = tan(x) / cos²(x) and that d/dx tan(x) = 1 / cos²(x).

Let u := xtan(x), then du = x / cos²(x) + tan(x) and

dv := 1 / cos²(x), then v = tan(x).

Using integration by parts we get

∫ xtan(x) / cos²(x) dx = ∫ udv = uv - ∫ vdu = xtan²(x) - ∫ xtan(x) / cos² + tan²(x) dx = xtan²(x) - ∫ xtan(x) / cos²(x) dx - ∫ tan²(x) dx. Taking the original integral to the left hand side we find

2∫ xtan(x) / cos²(x) = xtan²(x) - ∫ tan²(x) dx = xtan²(x) - tan(x) + x.

Hence we find

∫ xtan(x) / cos²(x) = [xtan²(x) - tan(x) + x] / 2.

Note that for ∫ tan²(x) dx we used the fact that d/dx tan(x) = 1 + tan²(x), hence tan(x) = ∫ 1 + tan²(x) dx, which yields ∫ tan²(x) dx = tan(x) - x.