# Help: I am having trouble solving this integral

$\large \int_{-a}^a \ln \left( \dfrac{a + \sqrt{a^2+ x^2}}{-a + \sqrt{a^2 + x^2}} \right) \, dx = \, ?$

Note by Anurag Pandey
1 year, 10 months ago

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Let $$x = a \tan\theta \Rightarrow dx = a\sec^2 \theta \, d\theta , \sqrt{a^2 + x^2} = a | \sec \theta |$$.

Since the integrand is an even function, then it can be simplified to

$\begin{eqnarray} I: &=& 2 \int_0^a \ln \left( \dfrac{a + \sqrt{a^2+ x^2}}{-a + \sqrt{a^2 + x^2}} \right) \, dx \\ &=& 2\int_0^{\pi /4} \ln \left( \dfrac{a + a |\sec \theta|}{-a + a |\sec \theta|} \right) \cdot a\sec^2 \theta \, d\theta \\ &=& 2 a \int_0^{\pi/4} \ln \left( \dfrac{1 + |\sec \theta|}{-1 + |\sec \theta|} \right) \sec^2 \theta \, d\theta \\ &=& 2 a \int_0^{\pi/4} \ln \left( \dfrac{1 + \sec \theta}{-1 + \sec \theta} \right) \sec^2 \theta \, d\theta \\ &=& 2 a \int_0^{\pi/4} \ln \left( \dfrac{\cos \theta + 1}{1 - \cos \theta} \right) \sec^2 \theta \, d\theta \\ &=& 2 a \int_0^{\pi/4} \ln \left( \dfrac{2\cos^2 \frac x2}{2\sin^2 \frac x2} \right) \sec^2 \theta \, d\theta \\ &=& 2 a \int_0^{\pi/4} \ln \left( \cot \dfrac \theta2 \right)^2 \sec^2 \theta \, d\theta \\ \dfrac I{4a} &=& \int_0^{\pi/4} \underbrace{ \ln \cot \frac \theta 2}_{=u} \cdot \underbrace{\sec^2 \theta \, d\theta}_{dv} \qquad, \qquad \text{ Integrate by parts} \\ &=& \left [ \ln \left( \cot \dfrac \theta2 \right) \tan \theta \right]_{\theta\to 0}^{\theta \to \pi /4} - \int_0^{\pi/4} \dfrac d{d\theta} \left ( \ln \cot \dfrac \theta 2 \right) \cdot \tan \theta \, d\theta \\ &=& \left [\left(\tan \dfrac\pi 4\right) \ln \left( \cot \dfrac\pi8\right) - \underbrace{\lim_{z\to0} \tan z \ln \cot \left( \dfrac z2 \right) }_{=0,\text{ See }(\bigstar) } \right ] - \int_0^{\pi/4} \dfrac{-\csc^2 \frac \theta2 \cdot \frac12}{\cot\frac\theta2} \cdot \tan \theta \, d\theta \\ &=& \left [ \ln \left( \cot \dfrac\pi8\right) - 0 \right] + \dfrac12 \int_0^{\pi/4} \dfrac{\csc^2 \frac \theta2 }{\cot\frac\theta2} \cdot \tan \theta \, d\theta \\ &=& \ln \left( \cot \dfrac\pi8\right) + \dfrac12 \int_0^{\pi/4} 2 \sec \theta \, d\theta \\ &=& \ln \left( \underbrace{\cot \dfrac\pi8}_{= \sqrt2+1, \text{ See }(\bigstar\bigstar)} \right) + \int_0^{\pi /4} \dfrac{\sec \theta (\sec\theta + \tan \theta)}{\sec\theta + \tan \theta} \, d\theta \\ &=& \ln (\sqrt 2 + 1) + \int_0^{\pi /4} \dfrac {d}{d\theta} \ln | \sec \theta + \tan \theta | \, d\theta \\ &=& \ln (\sqrt 2 + 1) + \left [ \ln | \sec \theta + \tan \theta | \right]_0^{\pi /4} \\ &=& \ln (\sqrt 2 + 1) + \left [ \ln \left(\sec \dfrac\pi4 + \tan \dfrac\pi4 \right) - \ln (\sec 0+ \tan 0) \right ] \\ &=& \ln (\sqrt 2 + 1) + \ln (1 + \sqrt2) = 2 \ln (1 + \sqrt2) \\ I &=& \boxed{8a \ln (1 + \sqrt2)} \approx 7.050988a \; . \end{eqnarray}$

Notes:

For $$(\bigstar)$$: We can rewrite the limit as $$\displaystyle \lim_{z\to0} \dfrac{ \ln\left(\cot \frac z2 \right)}{\frac1{\cot z}}$$, then apply L'Hôpital's rule to obtain a value of 0.

For $$(\bigstar\bigstar)$$: To prove that $$\cot \dfrac\pi8 = \sqrt2 +1$$ is equivalent to proving that $$\tan \dfrac\pi8 = \sqrt2 - 1$$. This can be shown by applying the double angle identity $$\tan (2A) = \dfrac{2\tan A}{1- \tan^2 A}$$, where $$A = \dfrac\pi8$$ followed by the quadratic formula.

- 1 year, 10 months ago

Why is it $+1/2 \int \sec \theta$? Shouldn't it be sec theta alone?

- 1 year, 10 months ago

Thanks for spotting my mistake. I've made the necessary changes.

- 1 year, 10 months ago

Thanks a lot. !

- 1 year, 10 months ago

Its one of the best Latex i've seen till now(more than 3-4 years) on brilliant

- 1 year, 10 months ago