Use integration by parts to evaluate the integral

Question

a. ∫cos^-1xdx

I did u=cos^-1x and dv=dx, so du=-1/(sqrt(1-x^2))dx and v=x. I'm not sure if this was right, but I proceeded anyway and got ∫cos^-1xdx=cos^-1x(x)=∫x(-1/(sqrt(1-x^2))dx. But I'm not sure if this was correct.

b. ∫e^-xcos(2x)dx

Would u=e^-x? And dv=cos(2x)dx for this question?

Answer 1

a. ∫Cos^-1xdx

Let u = Cos^-1x and dv = dx. So, du = (-1 / √(1 - x²))dx and v = x.

∫Cos^-1xdx = xCos^-1x + ∫ [x / √(1 - x²)]dx

Let w = 1-x². So, dw = -2xdx. xdx = (-1/2)dw

∫Cos^-1xdx = xCos^-1x -(1/2)∫w^-1/2dw = xCos^-1x - √w + C = xCos^-1x - √(1 - x²) + C

b. ∫e^-xcosxdx

Let u = e^-x and dv = cosxdx. So, du = -e^-xdx and v = ∫dv = sinx

∫e^-xcosxdx = e^-xsinx + ∫e^-xsinxdx

Evaluate ∫e^-xsinxdx by parts with u = e^-x and then combine like terms.