1. ## integration by parts

Hi the question is evaluate the integral
integral sign (S) cos^-1 (2x) dx using integration by parts

SO I let t = 2x so dt = 2 dx so dx= dt/2

So the original equation becomes S cos^-1 t (dt/2)

Then I let u = cos^-1 t so du = -dt/(1-t^2))^(1/2)

then dv = dt so v = t

So the original equation is now (1/2) S cos^-1 t dt
= (1/2) ((u cos^-1 u ) - S u (-du/sqrt(1-u^2)))
= (1/2) (u cos-1u - (1/2)u^2 (-du/sqrt (1-u^2)))

Is there any way to simplify this further or do I just substitute 2x and 2 dx back into the equation?

Thanks
Calculus beginner

2. $\displaystyle u = \arccos(2x)$

$\displaystyle dv = dx$

$\displaystyle du = -\frac{2}{\sqrt{1 - (2x)^2}} \, dx$

$\displaystyle v = x$

$\displaystyle \int \arccos(2x) \, dx = x\arccos(2x) + \int \frac{2x}{\sqrt{1 - (2x)^2}} \, dx$

now use the substitution $\displaystyle t = 2x$ for the last integral