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What would be the best way to approach the integral

\begin{equation*} \int_{0}^{\pi/2} \ln(a^2\sin^2x+b^2\cos^2x)dx \end{equation*}

I've tried imitating the method here to no avail. Would appreciate any help! Thank you!

Rajat
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David
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  • Have you tried taking partials with $a$ and $b$, looks pretty promising, not sure though. – Ahmed S. Attaalla May 28 '17 at 07:03
  • I haven't tried it and I'm unfamiliar with the method! Is that method differentiating under the integral sign? – David May 28 '17 at 07:09
  • Yeah that's what I was thinking, in-capturing some information about the partial derivatives, then using that for $f(a,b)=\int_{0}^{\pi/2} \ln(a^2\sin^2x+b^2\cos^2x)dx$ we have $f(1,1)=0$. – Ahmed S. Attaalla May 28 '17 at 07:12

1 Answers1

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I'll assume $a, b > 0$.

If your integral is $f(a,b)$, then $f(b,b) = \pi \ln b$ and $$\dfrac{\partial f}{\partial a} = \int_0^{\pi/2} \frac{2 a \sin^2 x}{a^2 \sin^2 x + b^2 \cos^2 x}\; dx = \frac{\pi}{a+b}$$ Conclude $$ f(a,b) = \pi \ln((a+b)/2) $$

Robert Israel
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