Integrating functions with algebraic and trigonometric parts. $\int\frac{x}{\sec x + 1}dx$

Question

$$\int\frac{x}{\sec x + 1}dx$$ How to perform this integration? I tried simplifying it to $$\frac{x \cos x}{1 + \cos x}$$ but after that integration by parts is not useful.

Answer 1

HINT:

$$\dfrac x{\sec x+1}=\dfrac{x(1+\cos x-1)}{1+\cos x}=x-\dfrac{x\sec^2\dfrac x2}2$$

For the second integral, integrate by parts

Answer 2

HINT: Integrate by parts: let $u = x, dv =\frac{1}{1+\sec \left(x\right)}dx$ then $du = dx, v = x-\tan \left(x/2\right)$.

$$∫\frac{x}{\sec x + 1}dx = uv - ∫ vdu = x^2-x\tan \left(x/2\right) -∫ x-\tan \left(x/2\right) dx$$