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Question How to solve the following integral? $$I = \int_{\pi/6}^{\pi/3} \dfrac{\sin x + \cos x}{\sqrt{\sin(2x)}}\,dx$$

Attempt Using the property of definite integrals and putting $\pi/6 + \pi/3 - x$ in place of $x$ did not help. So I tried simplifying the function and I got $[(\tan x)^{1/2} + (\cot x)^{1/2}]/2^{1/2}$, with which I don't know how to proceed?

Any helps appreciated.

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

$$\int{\frac{(\sin{x}+\cos{x})dx}{\sqrt{1-(\sin{x}-\cos{x})^2}}}$$

Let,

$$\sin{x}-\cos{x}=t$$ $$\frac{dt}{dx}=\sin{x}+\cos{x}$$

$$\int \frac{dt}{\sqrt{1-t^2}}$$

$$=\arcsin{t}=\arcsin{(\sin{x}-\cos{x})}$$

Putting limits,

$$=\arcsin{\frac{\sqrt{3}-1}{2}}-\arcsin{\frac{1-\sqrt{3}}{2}}$$

As a sidenote, I have always been taught to substitute $$\sin{2x}=1+(\sin{x}+\cos{x})^2$$ when there are $(\sin{x}-\cos{x})$ and $\sin{2x}$ in the same integral. $$\sin{2x}=(\sin{x}-\cos{x})^2-1$$ when there are $(\sin{x}+\cos{x})$ and $\sin{2x}$ in the same integral. Although exceptions may exist.

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  • $\begingroup$ All right, I retract my previous objection. $\endgroup$ – Jack D'Aurizio Sep 1 '18 at 16:04
  • $\begingroup$ @JackD'Aurizio, Also added a little sidenote. $\endgroup$ – prog_SAHIL Sep 1 '18 at 16:06
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The interval $\left(\frac{\pi}{6},\frac{\pi}{3}\right)$ is centered at $\frac{\pi}{4}$ and $\sin(x)+\cos(x)= \sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$, hence the given integral can be written as

$$ \int_{-\pi/12}^{\pi/12}\frac{\sqrt{2}\sin\left(x+\frac{\pi}{2}\right)}{\sqrt{\sin\left(2x+\frac{\pi}{2}\right)}}\,dx=\sqrt{2}\int_{-\pi/12}^{\pi/12}\frac{\cos(x)}{\sqrt{\cos(2x)}}\,dx=2\sqrt{2}\int_{0}^{\pi/12}\frac{\cos(x)}{\sqrt{\cos(2x)}}\,dx$$ or as $$ \sqrt{2}\int_{0}^{\pi/6}\frac{\cos(x/2)}{\sqrt{\cos x}}\,dx=\int_{0}^{\pi/6}\sqrt{\frac{1+\cos x}{\cos x}}\,dx=\int_{\sqrt{3}/2}^{1}\sqrt{\frac{1+x}{x}}\frac{dx}{\sqrt{1-x^2}} $$ or as $$ \int_{\sqrt{3}/2}^{1}\frac{dx}{\sqrt{x(1-x)}}=2\int_{\sqrt{\sqrt{3}/2}}^{1}\frac{du}{\sqrt{1-u^2}}=\pi-2\arcsin\sqrt{\frac{\sqrt{3}}{2}}=\boxed{\arccos(\sqrt{3}-1)}. $$

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$${\displaystyle\int}\dfrac{\sin\left(x\right)+\cos\left(x\right)}{\sqrt{\sin\left(2x\right)}}\,\mathrm{d}x$$ $$\sin\left(2x\right)=2\cos\left(x\right)\sin\left(x\right)$$ $$\sin\left(x\right)=\dfrac{\tan\left(x\right)}{\sec\left(x\right)}$$ $$\sec^2\left(x\right)=\tan^2\left(x\right)+1$$

$$={\displaystyle\int}\class{steps-node}{\cssId{steps-node-1}{\sec^2\left(x\right)}}\cdot\class{steps-node}{\cssId{steps-node-2}{\dfrac{\tan\left(x\right)+1}{\sqrt{2}\sqrt{\tan\left(x\right)}\left(\tan^2\left(x\right)+1\right)}}}\,\mathrm{d}x$$

Substitute $u=\tan\left(x\right)$ $$\class{steps-node}{\cssId{steps-node-3}{\dfrac{1}{\sqrt{2}}}}{\displaystyle\int}\dfrac{u+1}{\sqrt{u}\left(u^2+1\right)}\,\mathrm{d}u$$

Substitute $v=\sqrt{u}$

$$\class{steps-node}{\cssId{steps-node-4}{\sqrt 2}}{\displaystyle\int}\dfrac{v^2+1}{v^4+1}\,\mathrm{d}v$$

can you solve it using partial fraction decomposition

$${\sqrt 2}{\displaystyle\int}\dfrac{v^2+1}{\left(v^2-\sqrt{2}v+1\right)\left(v^2+\sqrt{2}v+1\right)}\,\mathrm{d}v$$

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  • $\begingroup$ Actually the tangent (half-angle) substitution does not seem to be the fastest way to tackle the given integral. $\endgroup$ – Jack D'Aurizio Sep 1 '18 at 15:25
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Set $u=\sin{x}-\cos{x}$ so $du = (\cos{x}+\sin{x})dx$ and $\sin{2x} =1-u^2$

Indefinite integral is then $$\int\frac{du}{(1-u^2)^{\frac{1}{2}}} = \sin^{-1}{u} =\sin^{-1}{(\sin{x}-\cos{x})}$$

Apply limits for answer of $ 2\sin^{-1} { (\frac{\sqrt{3}}{2} - \frac{1}{2}) } $

A bit of manipulation shows this is same as $\cos^{-1}{ (\sqrt{3}-1 ) }$

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Using $x=\tan^{-1}(t)$, the problem reduces to the computation of $$I=\frac{1}{\sqrt{2}}\int \frac{t+1}{ \sqrt{t} \left(t^2+1\right)}\,dt$$ which looks like the arctangent of something.

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