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I am trying to find the integral of $\dfrac{\sin(x)-\cos(x)}{\sqrt{\sin(2x)}}$.

I was trying to use the parametric equations to solve it, but it just turned into a much more complicated problem, moreover, this question seemed very specific and concrete.

Is there any intuition to solve a problem like this?

Sangchul Lee
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3b1b aimer
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3 Answers3

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Hint. Observe that $$(\sin x+\cos x)^2=\sin^2 x+2\sin x\cos x+\cos^2 x=1+\sin(2x),$$ so $$\sin(2x)=(\sin x+\cos x)^2-1.$$ Meanwhile, $$\frac{\mathrm{d}}{\mathrm{d}x}(\sin x+\cos x)=\cos x-\sin x=-(\sin x-\cos x).$$ Try the substitution method with $u:=\sin x+\cos x$. For more references, please see this post.

Bernard Pan
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We have that $$I=\int \frac{sin(x)-cos(x)}{\sqrt{sin(2x)}}dx=\int \frac{sin(x)-cos(x)}{\sqrt{2sin(x)cos(x)}}dx=\frac{1}{\sqrt 2}\int(\sqrt{tan(x)}-\frac{1}{\sqrt{tan(x)}})dx,$$ from there make the substitution $u=\sqrt{tan(x)}$.

Vaskara_GRek_O
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Letting $\sin x+\cos x=\cosh \theta$ yields $$I=-\int \frac{\sinh \theta }{\sqrt{\cosh^2 \theta -1}}d\theta=-\theta +C=-\cosh^{-1}(\sin x+\cos x)+C$$

Lai
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