\(\displaystyle{ \int{ \arccos x ~ dx } }\)

\(\displaystyle{ u = \arccos x \to du = \frac{-1}{\sqrt{1-x^2}} ~dx }\)

\(\displaystyle{ dv = dx \to v = x }\)

\(\displaystyle{ x\arccos x - \int{ \frac{-x}{\sqrt{1-x^2}} ~dx } }\)

Use integration by substitution. \(\displaystyle{ u=1-x^2 \to du = -2x~dx \to du/2 = -x~dx }\)
This \(u\) is different than the \(u\) used in integration by parts above.

\(\displaystyle{ x\arccos x - \int{ \frac{1}{u^{1/2}} \frac{du}{2} } }\)

\(\displaystyle{ x \arccos x - \frac{1}{2}\int{ u^{-1/2} ~ du } }\)

\(\displaystyle{ x \arccos x - \frac{1}{2}\frac{u^{1/2}}{1/2} + C }\)

\(\displaystyle{ x \arccos x - \sqrt{1-x^2} + C }\)

\(\displaystyle{ \int_1^3{ (4-x)^3 ~dx } }\)

\(\displaystyle{ u = 4-x \to du = -dx }\)

upper limit: \( x = 3 \to u = 4-3 = 1\)

lower limit: \( x = 1 \to u = 4-1 = 3 \)

\(\displaystyle{ \int_3^1{ u^3 (-du)} }\)

Switch the limits of integration to get rid of the negative sign.

\(\displaystyle{ \int_1^3{u^3~du} }\)

\(\displaystyle{ \left. \frac{u^4}{4} \right|_1^3 }\)

\(\displaystyle{ \frac{3^4}{4} - \frac{1^4}{4} }\)

\(\displaystyle{ \int_0^{\pi}{ \frac{\sin^3 x}{1 + \cos^2 x} ~dx } }\)

Drop the limits of integration and integrate the indefinite integral \(\displaystyle{ \int{ \frac{\sin^3 x}{1 + \cos^2 x} ~dx } }\)

\(\displaystyle{ \int{ \frac{\sin x(1-\cos^2 x)}{1 + \cos^2 x} ~dx } }\)

\(\displaystyle{ u = \cos x \to -du = \sin x ~dx }\)

\(\displaystyle{ -\int{ \frac{1-u^2}{1+u^2} ~du } }\)

\(\displaystyle{ \int{ \frac{u^2-1}{u^2+1} ~du } }\)

Do polynomial long division to get \(\displaystyle{ \int{ 1 - \frac{2}{u^2+1} ~du } }\)

\(\displaystyle{ u - 2\arctan u }\)

\(\displaystyle{ \cos x - 2\arctan(\cos x) }\)

Evaluate this result at the endpoints \(\displaystyle{ \left[ \cos x - 2\arctan(\cos x) \right]_0^{\pi} }\)

\(\displaystyle{ [\cos \pi - 2\arctan(\cos \pi)] - [ \cos 0 - 2 \arctan(\cos 0) ] }\)

\(\displaystyle{ -1 - 2\arctan(-1) - [1 - 2\arctan(1)] }\)

\(\displaystyle{ -1 - 2(-\pi/4) -1 + 2(\pi/4)}\)

\(\displaystyle{ -2+\pi }\)

\(\displaystyle{ \int{ 2x\frac{\sin^3 x}{1 + \cos^2 x} ~dx } }\)

Use integration by parts \(\displaystyle{ u = 2x \to du = 2dx }\)

\(\displaystyle{ dv = \frac{\sin^3 x}{1+\cos^2 x} ~dx }\)

\(\displaystyle{ v = \int{ \frac{\sin^3 x}{1+\cos^2 x} ~dx } }\)

\(v\) is the integral we did in the first part of the problem, so we have \( v = \cos x - 2\arctan(\cos x) \)

\(\displaystyle{ 2x[ \cos x - 2\arctan(\cos x) ] - 2 \int{ \cos x - 2\arctan(\cos x) ~dx }}\)

Now let's break the last integral into two.

\(\displaystyle{ 2\int{\cos x ~ dx } = 2\sin x }\)

When we evaluate this on the limits of integration, we get \( 2\sin \pi - 2\sin 0 = 0\)

For the second integral we have \(\displaystyle{ \int{ \arctan(\cos x) ~ dx } }\)

We are going to put the limits of integration back in since we will need them for the next step. So we have \(\displaystyle{ \int_0^{\pi}{ \arctan(\cos x) ~ dx } }\)

We will use technique 2 above to shift this integrand. \(\displaystyle{ \int_0^{\pi}{ \arctan(\cos x) ~ dx } = \int_0^{\pi}{ \arctan(\cos(\pi-x)) ~dx } }\)

\(\displaystyle{ \int_0^{\pi}{ \arctan(-\cos(x)) ~dx } }\)

\(\displaystyle{ \int_0^{\pi}{ -\arctan(\cos(x)) ~dx } }\)

So now we have \(\displaystyle{ \int_0^{\pi}{ \arctan(\cos x) ~ dx } = -\int_0^{\pi}{ \arctan(\cos x) ~ dx } }\)

If we add \(\displaystyle{ \int_0^{\pi}{ \arctan(\cos x) ~ dx } }\) to both sides, we have \(\displaystyle{ 2 \int_0^{\pi}{ \arctan(\cos x) ~ dx } = 0 }\)

So \(\displaystyle{ \int_0^{\pi}{ \arctan(\cos x) ~ dx } = 0 }\)

We are left with \(\displaystyle{ 2x[ \cos x - 2\arctan(\cos x) ] }\). This is the \(uv\) part of the integration by parts.

Evaluating this on the limits of integration gives us \(\displaystyle{ 2x[ \cos x - 2\arctan(\cos x) ]_0^{\pi} }\)

\(\displaystyle{ 2\pi[\cos \pi - 2\arctan(-1)] - 0 }\)

\(\displaystyle{ 2\pi[ -1 + \pi/2] }\)

\(\displaystyle{ \pi(\pi-2) }\)