Integration with substitution $u=\cos(x/2)$

Question

Please could you help me solve this integral

Find $$ \int \cos x \, \sqrt{1-\cos x} \, dx.$$

Hint: use the substitution $u=\cos(x/2)$.

Thanks.

Answer 1

You are given the substitution (also known as a change of variable) $u=\cos(x/2)$ to use.

Now using the half-angle identity,

$\cos(x)=\cos^2(x/2)-\sin^2(x/2)=2\cos^2(x/2)-1=2u^2-1$ (Eq. $1$)

so that

$1-\cos(x)=1-(2u^2-1)=2(1-u^2)$ (Eq. $2$)

From $u=\cos(x/2)$, we have

$\large \frac{du}{dx}=-\frac{1}{2}\sin(x/2)=-\frac{1}{2}\sqrt{1-u^2}$

leading to

$\large dx=-2\frac{du}{\sqrt{1-u^2}}$ (Eq. $3$)

Putting Equations $1$,$2$ and $3$ together we have

$\int \cos(x)\sqrt{1-\cos(x)}dx=\int(2u^2-1)\sqrt{2}\sqrt{1-u^2}\frac{-2}{\sqrt{1-u^2}}du=-2\sqrt{2}\int(2u^2-1)du$

Note that the $\sqrt{1-u^2}$ terms cancel, simplifying the integral

$-2\sqrt{2}\int(2u^2-1)du=-2\sqrt{2}[\frac{2u^3}{3}-u]+C$

where $C$ is a constant.

The last step is to replace $u$ by $\cos(x/2)$, leading to the final result being

$ \int \cos(x)\sqrt{1-\cos(x)}dx= 2\sqrt{2}\cos(x/2)-\frac{4\sqrt{2}}{3}\cos^3(x/2)+C$

Answer 2

Use the half angle identity: $1-\cos(x)=2\sin^2(\frac{x}{2})$ $$...=\int \sqrt{2} \sqrt{\sin^2\left(\frac{x}{2}\right)}\cos(x)dx$$ Take out the constant, and simplify power: $$...=\sqrt{2} \int \sin\left(\frac{x}{2}\right)\cos(x)dx$$ Use trig identity: $\sin(a)\cos(b)=\frac{1}{2}(\sin(a-b)-\sin(a+b)) $ $$...=\sqrt{2} \int \left(\sin\left(\frac{3x}{2}\right)-\sin\left(\frac{x}{2}\right)\right)dx$$ $$...$$ Skipped some steps: $$...=\sqrt{2}\cos\left(\frac{x}{2}\right)-\frac{\sqrt{2}}{3}\cos\left(\frac{x}{2}\right) + C$$

PS: I'm not 100% about the solution, but you may work this out by yourself to check!