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

I think there's no closed form antiderivative of it. I tried WolframAlpha but it didn't help much.

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  • $\begingroup$ where exactly is the square? $\endgroup$ – Dr. Sonnhard Graubner May 26 '17 at 11:51
  • $\begingroup$ In the last form, the value of the integral is just $\frac{1}{3}$. $\endgroup$ – Jack D'Aurizio May 26 '17 at 11:53
  • $\begingroup$ I was having some problems with Mathjax. Now there is no problem in the question. $\endgroup$ – Partha Sarker May 26 '17 at 11:53
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{}$.

\begin{align} &\int_{0}^{\pi/2}{\cos\pars{x} \over \bracks{1 + \root{\sin\pars{2x}}}^{2}} \,\dd x \,\,\,\,\,{\large\substack{x\ \mapsto\ x\ +\ \pi/4 \\[1mm] =}}\,\,\, \int_{-\pi/4}^{\pi/4}{\cos\pars{x}\cos\pars{\pi/4} - \sin\pars{x}\sin\pars{\pi/4} \over \bracks{1 + \root{\cos\pars{2x}}}^{2}} \,\,\dd x \\[5mm] = &\ \root{2}\int_{0}^{\pi/4}{\cos\pars{x} \over \bracks{1 + \root{1 -2\sin^{2}\pars{x}}}^{2}}\,\dd x \,\,\,\stackrel{\sin\pars{x}\ \mapsto\ x}{=}\,\,\, \root{2}\int_{0}^{\root{2}/2}{\dd x \over \bracks{1 + \root{1 - 2x^{2}}}^{2}} \end{align}

With Euler Substituion $\ds{t \equiv \root{1 - 2x^{2}} - \root{2}x\ic}$ :

\begin{align} &\int_{0}^{\pi/2}{\cos\pars{x} \over \bracks{1 + \root{\sin\pars{2x}}}^{2}} \,\dd x = 2\ic\int_{1}^{-\ic}{t^{2} + 1 \over \pars{t + 1}^{4}}\,\dd t = 2\ic\int_{2}^{1 - \ic} \pars{{1 \over t^{2}} - {2 \over t^{3}} + {2 \over t^{4}}}\,\dd t = \bbx{1 \over 3} \end{align}

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HINT:

Use $\displaystyle I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$

$$\displaystyle I+I=\int_a^b\{f(x)+f(a+b-x)\}\ dx$$

Now as $\displaystyle\int(\cos x+\sin x)dx=\sin x-\cos x, $ set $\sin x-\cos x=u\implies1-\sin2x=u^2$

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By setting $x=\frac{t}{2}$, then exploiting symmetry: $$ I=\int_{0}^{\pi/2}\frac{\cos(x)}{\left(1+\sqrt{\sin(2x)}\right)^2} = \frac{1}{2\sqrt{2}}\int_{0}^{\pi/2}\frac{\sqrt{1+\cos(x)}+\sqrt{1-\cos(x)}}{\left(1+\sqrt{\sin(x)}\right)^2}\,dx $$ and the last integral can be written in the following form: $$ I = \frac{1}{2}\int_{0}^{\pi/2}\frac{\sqrt{1+\sin x}}{1+\sin x+2\sqrt{\sin x}}\,dx=\frac{1}{2}\int_{0}^{1}\frac{\sqrt{1+x}}{\sqrt{1-x^2}}\cdot\frac{dx}{(1+\sqrt{x})^2} $$ or the following one: $$ I = \frac{1}{2}\int_{0}^{1}\frac{dx}{\sqrt{1-x}(1+\sqrt{x})^2} = \frac{1}{2}\int_{0}^{1}\frac{dx}{\sqrt{x}(1+\sqrt{1-x})^2}=\int_{0}^{1}\frac{dx}{\left(1+\sqrt{1-x^2}\right)^2}$$ that also equals $$ I = \int_{0}^{\pi/2}\frac{\cos t}{(1+\cos t)^2}\,dt = \int_{0}^{2}\frac{\frac{1}{4}-\frac{u^4}{64}}{1+\frac{u^2}{4}}\,du = \frac{1}{16}\int_{0}^{2}\left(4-u^2\right)\,du = \color{red}{\frac{1}{3}}$$ by Weierstrass substitution $t=2\arctan\frac{u}{2}$.

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  • $\begingroup$ I don't understand how you get (1+cosx)^1/2 + 1-cosx)^1/2 from the first line $\endgroup$ – Partha Sarker May 30 '17 at 23:52
  • $\begingroup$ @Mockingbird360: I replace $x$ with $\frac{z}{2}$, then exploit $\sin(\pi-z)=\sin(z)$ and the bisection formulas for $\sin$ and $\cos$. $\endgroup$ – Jack D'Aurizio May 31 '17 at 1:06

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