Home » Uncategorized » An integral with roots

An integral with roots

Prove that

    \[\int_{-1}^{1} \frac{\mathrm{d}x}{\sqrt{1+x} + \sqrt{1-x} + 2} = 4\sqrt{2} - \pi - 2\]

Solution

First of all we note that

    \[\left ( \sqrt{1+x} + \sqrt{1-x} \right )^2 = 2\left ( 1+ \sqrt{1-x^2} \right )\]

Hence,

    \begin{align*} \int_{-1}^{1} \frac{\mathrm{d}x}{\sqrt{1+x} + \sqrt{1-x} + 2} &= \int_{-1}^{1} \frac{\mathrm{d}x}{\sqrt{2\left (1+\sqrt{1-x^2} \right )}+2} \\ &\!\!\!\!\!\overset{x = \sin u}{=\! =\! =\! =\! =\!} \int_{-\pi/2}^{\pi/2} \frac{\cos u}{\sqrt{2\left ( 1+\cos u \right )}+2}\, \mathrm{d}u\\ &=\int_{-\pi/2}^{\pi/2} \frac{\cos u}{\sqrt{4 \cos^2 \frac{u}{2}}+2} \, \mathrm{d}u \\ &= \int_{-\pi/2}^{\pi/2} \frac{\cos u}{2 \cos \frac{u}{2} + 2} \, \mathrm{d} u \\ &= \frac{1}{2} \int_{-\pi/2}^{\pi/2} \frac{\cos u}{1+ \cos \frac{u}{2}} \, \mathrm{d}u \\ &= \frac{1}{4} \int_{-\pi/2}^{\pi/2} \frac{\cos u}{\cos^2 \frac{u}{4}} \, \mathrm{d}u \\ &= \cancelto{0}{\frac{1}{4} \left [ 4 \tan \frac{u}{4} \cos u \right ]_{-\pi/2}^{\pi/2}} +\\ & \quad \quad \quad +  \int_{-\pi/2}^{\pi/2} \tan \frac{u}{4} \sin u \, \mathrm{d}u \end{align*}

To finish things off we evaluate the final integral. For example:

    \begin{align*} \int_{-\pi/2}^{\pi/2} \tan \frac{u}{4} \sin u \, \mathrm{d}u &\overset{y=u/4}{=\! =\! =\! =\!} 4 \int_{-\pi/8}^{\pi/8} \tan y \sin 4y \, \mathrm{d}y\\ &=4 \int_{-\pi/8}^{\pi/8} \tan y \left ( 4 \sin y \cos^3 y - 4 \sin^3 y \cos y \right )\, \mathrm{d}y \\ &=4 \int_{-\pi/8}^{\pi/8} \left ( 4 \sin^2 y \cos^2 y - 4 \sin^4 y \right )\, \mathrm{d}y \\ &= 16 \int_{-\pi/8}^{\pi/8} \left ( \sin y \cos y \right )^2 \, \mathrm{d} y - 16 \int_{-\pi/8}^{\pi/8} \sin^4 y \, \mathrm{d}y \\ &= \left (\frac{\pi}{2} - 1 \right ) + \left (-1 + 4 \sqrt{2} - \frac{3\pi}{2} \right ) \\ &= 4\sqrt{2} - \pi - 2 \end{align*}

in view of the identities

    \[\sin^4 x = \frac{1}{8} \left ( 3 + \cos 4x - 4 \cos 2x \right )\]

and

    \begin{align*} \int_{-\pi/8}^{\pi/8} \sin^2 x \cos^2 x \, \mathrm{d}x &= \int_{-\pi/8}^{\pi/8} \left ( \sin x \cos x \right )^2 \, \mathrm{d} x\\ &=\int_{-\pi/8}^{\pi/8} \left ( \frac{\sin 2x}{2} \right )^2 \, \mathrm{d}x\\ &= \frac{1}{4} \int_{-\pi/8}^{\pi/8} \sin^2 2x \, \mathrm{d}x\\ &=\frac{1}{8}\int_{-\pi/8}^{\pi/8} \left ( 1- \cos 4x \right ) \, \mathrm{d}x \\ &= \frac{1}{8} \left ( \frac{\pi}{4} - \frac{1}{2} \right ) \\ &= \frac{\pi}{32} - \frac{1}{16} \end{align*}

Read more

Leave a comment

Who is Tolaso?

Find out more at his Encyclopedia Page.

Donate to Tolaso Network