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In the following integration:

$$\int \frac{dx}{\sin{x}\sqrt{\sin(2x+\alpha)}}$$

My attempt is:

I substituted $\sin(2x+\alpha) = t$.

And got $2\cos(2x+\alpha)~dx =dt$.

But after that got stuck.

Can somebody help me.

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    where did you get this integral – 2017-01-07
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    [WolframAlpha](http://wolframalpha.com/input/?i=int+1%2F%28sin%28x%29*sqrt%28sin%28x%29%2Ba%29%29+dx&x=0&y=0) yields an ugly result. – 2017-01-07
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    Wolfram is not having this one... https://www.wolframalpha.com/input/?i=integral+of+1%2F(sin(x)sqrt(sin(2x%2Ba)))+dx – 2017-01-07
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    This is in my book – 2017-01-07
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    No textbook would give an example as challenging as this on purpose – 2017-01-07
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    What is the book? I've seen multiple users posting screenshots like this (there was a rash of them involving ridiculous limits), that clearly have a watermark and, presumably, are from the same book. – 2017-01-07
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    Noticed that as well pjs36... – 2017-01-07
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    The answer given in the book is http://i.imgur.com/oYu3Y4S.jpg – 2017-01-07
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    @pjs36 that book is available online thats why – 2017-01-07
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    What is the book's name? – 2017-01-07
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    use the addition formulas to simplify the square root term – 2017-01-07
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    @Dr.SonnhardGraubner but in addition formula there is no squared term then how can we simplify it – 2017-01-07
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    i meant $$\sin(2x+\alpha)=\sin(2x)\cos(\alpha)+\cos(2x)\sin(\alpha)$$ – 2017-01-07
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    I don't trust mathematics texts not written with LaTeX... – 2017-01-07
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    Beautiful formatting ^^! – 2017-01-07
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    Maple V (yes, I'm a luddite) gives an answer if I set $\alpha = 1$, but it's more than 4 pages of square roots of cosines and such. And I spot several occurrences of EllipticPi and and EllipticF in there. This ain't your grandma's integral. I'd set $\alpha=0$ and then you'll have a nice, challenging exercise. – 2017-01-07
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    @Dr.SonnhardGraubner this is what I have done in my attempt – 2017-01-08
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    Which book did you encounter this integral from? – 2017-02-01
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    @Dr.SonnhardGraubner Please reconsider this question – 2017-02-01

2 Answers 2

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$\sin(2x+A)=\sin2x\cos A+\cos2x\sin A$

$=\sin^2x\{2\cos A\cot x+(\cot^2x-1)\sin A\}$

Put $-\cot x=v$ to find $$I=\int\dfrac{dx}{\sin x\sqrt{\sin(2x+A)}}=(\text{sign of} \sin x)\int\dfrac{dv}{\sqrt{(v^2-1)\sin A-2v\cos A}}$$

Case$\#1:$ For $\sin A>0,$ $$\sqrt{\sin A}\cdot I=(\text{sign of} \sin x)\int\dfrac{dv}{\sqrt{(v^2-1)-2v\cot A}}$$

Now $\displaystyle\int\dfrac{dv}{\sqrt{(v^2-1)-2v\cot A}}=\int\dfrac{dv}{\sqrt{(v-\cot A)^2-\csc^2A}}$

Case$\#2:$ For $\sin A<0,$ $$\sqrt{-\sin A}\cdot I=(\text{sign of} \sin x)\int\dfrac{dv}{\sqrt{-(v^2-1)+2v\cot A}}$$

Now $\displaystyle\int\dfrac{dv}{\sqrt{1-v^2+2v\cot A}}=\int\dfrac{dv}{\sqrt{\csc^2A-(v-\cot A)^2}}$

Can yo take it from here?

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    Thanks for your answer.But i didnot understand why you broke it down into cases.please clarify a bit. – 2017-02-02
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    @navinstudent, As we are dealing with real Calculus, we need to be careful about the sign under radical i.e.$\sqrt{}$. First of all, $$\sqrt{\sin^2x\{2\cos A\cot x+(\cot^2x-1)\sin A\}}=\sqrt{\sin^2x}\cdot\sqrt{2\cos A\cot x+(\cot^2x-1)\sin A}$$ Now $\sqrt{\sin^2x}=|\sin x|=$sign of$(\sin x)\cdot\sin x$ – 2017-02-02
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    @navinstudent, Again, $2\cos A\cot x+(\cot^2x-1)\sin A=\sin x\{\cot^2x-1+2\cot A\cot x\}$ If $\sin A>0,$ $$\sqrt{2\cos A\cot x+(\cot^2x-1)\sin A}=\sqrt{\sin A}\cdot\sqrt{\cot^2x-1+2\cot A\cot x}$$ and if $\sin A<0,$ $$\sqrt{2\cos A\cot x+(\cot^2x-1)\sin A}=\sqrt{-\sin A}\cdot\sqrt{-(\cot^2x-1)-2\cot A\cot x}$$ – 2017-02-02
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    Excellent i used to neglect the signs often.Thanks again. – 2017-02-02
2

$$\int \frac{dx}{\sin x \sqrt{\sin 2x \cos a + \cos 2x \sin a}}$$ which is same as $$\int \frac{dx}{\sin x \sqrt{\sin 2x} \sqrt{ \cos a + \cot 2x \sin a}}$$ Now let $t=\cos a+\cot 2x \sin a$.Then $dt=-2cosec ^2 2x \sin a dx$.So that i could rewrite the integral as $$-\frac{√2}{\sin a} \int \frac{\sqrt{\sin x}(\cos x)^{3/2} dt}{√t}$$ I then tried to go in for integration by parts but that seemed too lengthy and cumbersome.