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My textbook has the following problem:

$$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \ dx$$

Using the trigonometric identity: $$\sin 2x = 2 \sin x \cos x$$

the integration simplifies to:

$$\frac{1}{\sqrt{2}} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) \ dx$$

I'm stuck here.

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  • $\begingroup$ One method is to find the indefinite integration...but that is inelegant imo $\endgroup$ Commented May 20 at 3:52

2 Answers 2

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\begin{align} &\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x +\cos x}{\sqrt{\sin 2x}} \ dx\\ =&\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d(\sin x -\cos x)}{\sqrt{1-(\sin x -\cos x)^2}} \\ =& \ \arcsin(\sin x -\cos x)\bigg|_{\frac{\pi}{6}}^{\frac{\pi}{3}} =2\arcsin\frac{\sqrt3-1}2 \end{align}

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    $\begingroup$ (+1) The solution can be further simplified to $\arccos(\sqrt3-1)$ or $\arctan\frac{\sqrt[4]3}{\sqrt2}$ $\endgroup$
    – user170231
    Commented May 22 at 18:01
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Not quite as fast, but we can simplify the numerator first by substituting $x\to x+\dfrac\pi4$:

$$\begin{align*} I &= \int_\tfrac\pi6^\tfrac\pi3 \frac{\sin x+\cos x}{\sqrt{\sin(2x)}} \, dx \\ &= \int_{-\tfrac\pi{12}}^\tfrac\pi{12} \frac{\sqrt2\,\cos x}{\sqrt{\cos(2x)}} \, dx \\ &= 2\sqrt2 \int_0^\tfrac\pi{12} \frac{d(\sin x)}{\sqrt{1-2\sin^2x}} \end{align*}$$

Easy enough, but we can further simplify the integral by substituting $\sin x=\dfrac1{\sqrt2}\sin y$:

$$I = 2\sqrt2 \int_0^{\arcsin\tfrac{\sqrt3-1}2} \frac{\cos y}{\sqrt2 \sqrt{1-\sin^2y}} \, dy = 2\arcsin\frac{\sqrt3-1}2 = \boxed{\arccos\left(\sqrt3-1\right)}$$

This follows from the double angle identity:

$$\sin\dfrac\pi{12}=\sqrt{\dfrac{1-\cos\frac\pi6}2} \\ \arcsin x=\dfrac12\arccos\left(1-2x^2\right)$$

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