How to integrate $\int \frac{\cos x}{\sqrt{\sin2x}} \,dx$?

Question

How to integrate $$\int \frac{\cos x}{\sqrt{\sin2x}} \,dx$$ ?

I have: $$\int \frac{\cos x}{\sqrt{\sin2x}} \,dx = \int \frac{\cos x}{\sqrt{2\sin x\cos x}} \,dx = \frac{1}{\sqrt2}\int \frac{\cos x}{\sqrt{\sin x}\sqrt{\cos x}} \,dx = \frac{1}{\sqrt2}\int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} \,dx = \frac{1}{\sqrt2}\int \sqrt{\frac{\cos x}{\sin x}} \,dx = \frac{1}{\sqrt2}\int \sqrt{\cot x} \,dx \\ t = \sqrt{\cot x} \implies x = \cot^{-1} t^2 \implies \,dx = -\frac{2t\,dt}{1 + t^4}$$

so I have: $$-\sqrt2 \int \frac{t^2 \,dt}{1 + t^4}$$

I tried partial integration on that but it just gets more complicated. I also tried the substitution $t = \tan \frac{x}{2}$ on this one: $\frac{1}{\sqrt2}\int \sqrt{\frac{\cos x}{\sin x}} \,dx$

$$= \frac{1}{\sqrt2}\int \sqrt{\frac{\frac{1 - t^2}{1 + t^2}}{\frac{2t}{1+t^2}}} \frac{2\,dt}{1+t^2} = \frac{1}{\sqrt2}\int \sqrt{\frac{1 - t^2}{2t}} \frac{2\,dt}{1+t^2} = \int \sqrt{\frac{1 - t^2}{t}} \frac{\,dt}{1+t^2}$$

... which doesn't look very promising.

Any hints are appreciated!

Answer 1

Having it the form of one polynomial divided by another should suggest one method, ugly though it may be: partial fractions. We have $1+t^4=1+2t^2+t^4-2t^2=(t^2+1-t\sqrt2)(t^2+1+t\sqrt2)$

Answer 2

I have a half answer (exactly as you did).

Put $\sin 2x=2\sin x\cos x$. Then you integral is:

$$\int\dfrac{\cos x}{\sqrt{2\sin x }\sqrt{\cos x}}dx=\int\dfrac{\sqrt{\cos x}}{\sqrt{2\sin x }}dx=\dfrac{\sqrt{2}}{2}\int\sqrt{\cot x}\,dx.$$

For the second half of the answer, see this

Answer 3

In fact you have $$\frac{x^2}{x^4+1}=\frac{1}{2\sqrt{2}}( \frac{x+1}{x^2-\sqrt{2}{x+1}}-\frac{x+1}{x^2+\sqrt{2}{x+1}})$$ now you can continue to integrate and get some log and arctan terms.

Answer 4

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{}$ \begin{align} &\int{\cos\pars{x} \over \root{\sin\pars{2x}}}\,\dd x =\pm\,{1 \over \root{2}}\ \overbrace{\int{\cos^{1/2}\pars{x} \over \sin^{1/2}\pars{x}}\,\dd x} ^{\ds{\mbox{Set}\ t \equiv \sin\pars{x}}}\ =\ \pm\,{\root{2} \over 2}\ \overbrace{\int t^{-1/2}\pars{1 - t^{2}}^{-1/4}\,\dd t}^{\ds{t \equiv y^{1/2}}} \\[3mm]&=\pm\,{\root{2} \over 4}\int y^{-3/4}\pars{1 - y}^{-1/4}\,\dd y \\[5mm]&\mbox{which can be related to a Beta Function for some 'nice' limits.} \end{align}

Answer 5

Using the identities $$ \sin 2 x =(\cos x+\sin x)^2-1=1-(\cos x-\sin x)^2 $$ We can split $\cos x$ in the numerator as $$ \begin{aligned} \int \frac{\cos x}{\sqrt{\sin 2 x}} d x=&\frac{1}{2} \int \frac{(-\sin x+\cos x)-(-\sin x-\cos x)}{\sqrt{\sin 2 x}} d x \\ = & \frac{1}{2}\left[\int \frac{d(\cos x+\sin x)}{\sqrt{(\cos x+\sin x)^2-1}}-\int \frac{d(\cos x-\sin x)}{\sqrt{1-(\cos x-\sin x)^2}}\right] \\ = & \frac{1}{2}\left[\cosh^{-1} (\cos x+\sin x)-\sin ^{-1}(\cos x-\sin x)\right]+C \end{aligned} $$

Answer 6

$\displaystyle\int\frac{cosxdx}{\sqrt{sin2x}}=\frac{1}{\sqrt{2}}\int\sqrt{\frac{cosx}{sinx}}dx=\frac{1}{\sqrt{2}}\int\sqrt{\cot{x}}dx$

$\displaystyle z^{2}=\cot{x}\Rightarrow 2zdz=-(1+z^{2})dx$

$\displaystyle\frac{1}{\sqrt{2}}\int\frac{2z^{2}dz}{1+z^{2}}=\sqrt{2}\int(1-\frac{1} {1+z^{2}})dz=\sqrt{2}\left( z-\arctan{z} \right)+c$

$\displaystyle\int\frac{cosxdx}{\sqrt{sin2x}}=\sqrt{2}(\sqrt{\cot{x}}-\arctan\sqrt{\cot{x}})+c$