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I'm stuck with the following

$$ \int_0^{2\pi} \cos^2(x)\sqrt{1+\cos(x)}\,dx. $$

I came out of $$ \int_0^{2\pi} \cos(x)\sqrt{1+\cos(x)}\,dx $$ integrating by parts, knowing that $$ \int \sqrt{1+\cos(x)}\,dx=\text{sgn}(\sin(x))2\sqrt{1-\cos(x)}+C $$ and thus integrating $\int_0^{2\pi} \cos(x)\sqrt{1+\cos(x)}\,dx$ first on $[0,\pi]$ and then on $[\pi,2\pi]$. I tried the same strategy (integration by parts) with the first integral, but it didn't work. I am also not sure this $$ \int \sqrt{1+\cos(x)}\,dx=\int \sqrt{1+\cos(x)}\frac{\sqrt{1-\cos(x)}}{\sqrt{1-\cos(x)}}dx=\int \frac{\sqrt{1-\cos^2(x)}}{{\sqrt{1-\cos(x)}}}dx=\int \frac{\sqrt{\sin^2(x)}}{{\sqrt{1-\cos(x)}}}dx=\int\frac{|\sin(x)|}{\sqrt{1-\cos(x)}}dx=\ldots $$ is the best way to come out of $\int \sqrt{1+\cos(x)}\,dx$. Thank you.

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    $\begingroup$ I can't tell exactly what the original question was, but perhaps the identity $\cos x = 2\cos^2 \tfrac 12 x - 1$ will come in handy. $\endgroup$ Jul 12, 2020 at 15:17
  • $\begingroup$ Try substituting $z=1-\cos(x)$ $\endgroup$
    – Integrand
    Jul 12, 2020 at 15:25

3 Answers 3

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Per the symmetry of the integrand with respect to $x=\pi$

\begin{align} & \int_0^{2\pi} \cos^2 x\sqrt{1+\cos x}dx\\ =& \ 2\sqrt2 \int_0^{\pi}\cos^2 x \cos\frac x2 dx\\ =& \ 4\sqrt2 \int_0^{\pi} \left(1-2\sin^2\frac x2\right)^2 \cos\frac x2\ dx\\ =&\ 4\sqrt2 \int_0^{\pi} (1-4\sin^2 \frac x2+4\sin^4 \frac x2)d(\sin \frac x2)\\ =&\ \frac{28\sqrt2}{15} \end{align}

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Use $$\int_{0}^{2a} f(x) dx=2 \int_{0}^{a} f(x) dx,~if~ f(2a-x)=f(x).$$ And $$\int_{0}^a f(x) dx=\int_{0}^{a} f(a-x) dx$$ Then $$I=\int_{0}^{2\pi} \cos^2 x \sqrt{1+\cos x}~dx=2\int_{0}^{\pi} \cos^2 x\sqrt{1-\cos x}~dx$$ $$I=2\int_{0}^{\pi} \frac{\cos^2 x \sin x}{\sqrt{1+\cos x}} dx$$ Let $\cos x=t \implies -\sin x~ dx =dt$ $$\implies I=2\int_{-1}^{1} \frac{t^2}{\sqrt{1+t}}=4\int_{0}^{\sqrt{2}} (u^2-1)^2 du=\frac{28\sqrt{2}}{15}.$$

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hint

With the substitution $$t=x-\pi$$

it becomes

$$\int_{-\pi}^\pi\cos^2(t)\sqrt{1-\cos(t)}dt=$$

$$2\int_0^{\pi}\Bigl(2\cos^2(\frac t2)-1\Bigr)^2\sqrt{2\sin^2(\frac t2)}dt=$$

$$2\sqrt{2}\int_0^\pi\Bigl(2\cos^2(\frac t2)-1\Bigr)^2\sin(\frac t2)dt$$

because $0\le \frac t2\le \frac{\pi}{2}$ and $\;\;\sin(\frac t2)\ge 0$.

Now, put $$u=\cos(\frac t2)$$ to get

$$4\sqrt{2}\int_0^1(2u^2-1)^2du$$ $$=4\sqrt{2}(\frac 45-\frac 43+1)=\frac{28\sqrt{2}}{15}$$

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