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By differentiating under the integral sign, evaluate I =…
Question
By differentiating under the integral sign, evaluate I = \int_0^{\pi/2} \frac{1}{\cos x}\,\ln\!\left(\frac{1+\tfrac12\cos x}{1-\tfrac12\cos x}\right)dx. It equals \frac{m}{n}\pi^2 with m/n in lowest terms. Find m+n.
✓ Verified answer: 7checked by our engine — not a guess
Step-by-step solution
Define I(a)=\int_0^{\pi/2}\frac{1}{\cos x}\ln\!\frac{1+a\cos x}{1-a\cos x}\,dx with parameter a.
Differentiate in a: I'(a)=\int_0^{\pi/2}\frac{1}{\cos x}\Big(\frac{\cos x}{1+a\cos x}+\frac{\cos x}{1-a\cos x}\Big)dx = \int_0^{\pi/2}\frac{2}{1-a^2\cos^2 x}\,dx.
Using \int_0^{\pi/2}\frac{dx}{1-a^2\cos^2 x}=\frac{\pi}{2\sqrt{1-a^2}}, we get I'(a)=\frac{\pi}{\sqrt{1-a^2}}.
Since I(0)=0, integrate: I(a)=\pi\arcsin a.
At a=1/2: I=\pi\arcsin(1/2)=\pi\cdot\frac{\pi}{6}=\frac{\pi^2}{6}.
So m=1, n=6 and m+n = 7.
Final answer7
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