Evaluate $$\int_0^{\pi/2}\log(a^2\sin^2x+b^2\cos^2x)dx$$ Basically, I have tried all the definite integrals formula I know, but nothing works. Please help me. I can't figure out actually how to handle this one.

Thanks in advance.

  • $\begingroup$ You may just set $x=\arctan t$ then perform an integration by parts to reduce the problem to the integration of a rational function. $\endgroup$ Jun 15 '20 at 17:41
  • $\begingroup$ Differentiate with respect to either $a$ or $b$ and use the tangent half angle substitution. Then integrate the result. $\endgroup$
    – Mark Viola
    Jun 15 '20 at 20:35

Use the known result $\int^{\pi}_{0} {\ln(1+r^2 -2r\cos t)}dt=0$ to obtain

\begin{align} & \int^{\frac{\pi}{2}}_{0}{\ln{\left(a^2\sin^2x+b^2\cos^2x\right)}dx}\\ = & \frac12\int^{\pi}_{0}{\ln{\left(\frac{a^2+b^2}2-\frac{a^2-b^2}2\cos t\right)}dt}\\ = & \frac12\int^{\pi}_{0} {\ln\left( \left(\frac{|a|+|b|}2\right)^2\cdot (1+r^2 -2r\cos t)\right)}dt ,\>\>\> r=\frac{|a|-|b|}{|a|+|b|}\\ = & \pi\ln\frac{|a|+|b|}2 \end{align}


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