3
$\begingroup$

$$\int_0^{2\pi}(t-\sin t){\sqrt{1-\cos t}} dt$$ I can notice that I have something of the form $$\int{f(x){\sqrt {f'(x)}}dx}$$ but I don't know anything that could simplify it

3 Answers 3

3

HINT:

Note that $\sqrt{1-\cos(x)}=\sqrt{2}|\sin(x/2)|$

The integral of $x\sin(x/2)$ can be evaluated using integration by parts, while the integral of $\sin(x)\sin(x/2)$ is facilitated using the addition angle formula.

2

Well, we have that

$$\mathcal{I}:=\int_0^{2\pi}\left(t-\sin\left(t\right)\right)\sqrt{1-\cos\left(t\right)}\space\text{d}t=$$ $$\int_0^{2\pi}t\sqrt{1-\cos\left(t\right)}\space\text{d}t-\int_0^{2\pi}\sin\left(t\right)\sqrt{1-\cos\left(t\right)}\space\text{d}t\tag1$$

Now, we get that:

  • Substiute $\text{u}=1-\cos\left(t\right)$: $$\int_0^{2\pi}\sin\left(t\right)\sqrt{1-\cos\left(t\right)}\space\text{d}t=\int_0^0\sqrt{\text{u}}\space\text{d}\text{u}=0\tag2$$
  • Substiute $\text{s}=\frac{t}{2}$: $$\int_0^{2\pi}t\sqrt{1-\cos\left(t\right)}\space\text{d}t=\sqrt{2}\int_0^{2\pi}t\sin\left(\frac{t}{2}\right)\space\text{d}t=4\sqrt{2}\int_0^\pi\text{s}\sin\left(\text{s}\right)\space\text{d}\text{s}\tag3$$

Now, using integration by parts:

$$4\sqrt{2}\int_0^\pi\text{s}\sin\left(\text{s}\right)\space\text{d}\text{s}=4\sqrt{2}\cdot\left(\pi+\int_0^\pi\cos\left(\text{s}\right)\space\text{d}\text{s}\right)=$$ $$4\sqrt{2}\cdot\left(\pi+\sin\left(\pi\right)-\sin\left(0\right)\right)=4\pi\sqrt{2}\tag4$$

So, we get that:

$$\mathcal{I}=4\pi\sqrt{2}$$

  • 0
    There is a problem with the substitution $u=1-\cos(t)$ on $[0,2\pi]$ since the mapping is not one-to-one. You need to break this into intervals over which this is legitimate ($[0,\pi]$ and$ [\pi,2\pi]$). Alternatively, exploit periodicity. The result is still $0$. – 2017-02-09
1

Another approach that saves you partial integration. First remark that: \begin{align} \int_0^{2\pi}f(t)\mathrm{d}t=\int_0^{2\pi}f(2\pi-t)\mathrm{d}t\end{align} In this particular case you will get: \begin{align} I=\int_0^{2\pi}(t-\sin(t))\sqrt[]{1-\cos(t)}\mathrm{d}t=\int_0^{2\pi}(2\pi-t+\sin(t))\sqrt[]{1-\cos(t)}\mathrm{d}t\end{align} Adding them to eachother yields, \begin{align} 2I=\int_0^{2\pi}2\pi\sqrt[]{1-\cos(t)}\mathrm{d}t\end{align} Now use the fact:$\sqrt[]{1-\cos(t)}=\sqrt[]{2}|\sin(t/2)|$ as pointed out by @dr-mv .