Help: I am having trouble solving this integral

aaln(a+a2+x2a+a2+x2)dx=? \large \int_{-a}^a \ln \left( \dfrac{a + \sqrt{a^2+ x^2}}{-a + \sqrt{a^2 + x^2}} \right) \, dx = \, ?

#Calculus

Note by Anurag Pandey
4 years, 9 months ago

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Comments

Let x=atanθdx=asec2θdθ,a2+x2=asecθx = 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

I:=20aln(a+a2+x2a+a2+x2)dx=20π/4ln(a+asecθa+asecθ)asec2θdθ=2a0π/4ln(1+secθ1+secθ)sec2θdθ=2a0π/4ln(1+secθ1+secθ)sec2θdθ=2a0π/4ln(cosθ+11cosθ)sec2θdθ=2a0π/4ln(2cos2x22sin2x2)sec2θdθ=2a0π/4ln(cotθ2)2sec2θdθI4a=0π/4lncotθ2=usec2θdθdv, Integrate by parts=[ln(cotθ2)tanθ]θ0θπ/40π/4ddθ(lncotθ2)tanθdθ=[(tanπ4)ln(cotπ8)limz0tanzlncot(z2)=0, See ()]0π/4csc2θ212cotθ2tanθdθ=[ln(cotπ8)0]+120π/4csc2θ2cotθ2tanθdθ=ln(cotπ8)+120π/42secθdθ=ln(cotπ8=2+1, See ())+0π/4secθ(secθ+tanθ)secθ+tanθdθ=ln(2+1)+0π/4ddθlnsecθ+tanθdθ=ln(2+1)+[lnsecθ+tanθ]0π/4=ln(2+1)+[ln(secπ4+tanπ4)ln(sec0+tan0)]=ln(2+1)+ln(1+2)=2ln(1+2)I=8aln(1+2)7.050988a  .\begin{aligned} 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{aligned}


Notes:

For ()(\bigstar): We can rewrite the limit as limz0ln(cotz2)1cotz \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 cotπ8=2+1 \cot \dfrac\pi8 = \sqrt2 +1 is equivalent to proving that tanπ8=21 \tan \dfrac\pi8 = \sqrt2 - 1. This can be shown by applying the double angle identity tan(2A)=2tanA1tan2A\tan (2A) = \dfrac{2\tan A}{1- \tan^2 A} , where A=π8A = \dfrac\pi8 followed by the quadratic formula.

Pi Han Goh - 4 years, 9 months ago

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

Prakhar Bindal - 4 years, 9 months ago

Thanks a lot. !

Anurag Pandey - 4 years, 9 months ago

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

Shaun Leong - 4 years, 9 months ago

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

Pi Han Goh - 4 years, 9 months ago
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