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Compute: $$I(y)=\int_0^{\frac {\pi}2} \ln(\cos^2x+y^2\sin^2x) dx$$

So$$I'(y)=\int_0^{\frac {\pi}2}\frac {2y\sin^2x}{\cos^2x+y^2\sin^2x}dx$$ Now what?

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$$I(y)=\int_0^{\frac {\pi}2} \ln(\cos^2x+y^2\sin^2x) dx\Rightarrow I'(y) =2y\int_0^\frac{\pi}{2}\frac{\sin^2 x}{\cos^2 x+y^2\sin^2 x}dx$$ $$=2y\int_0^\frac{\pi}{2} \frac{1}{\cot^2 x +y^2}dx\overset{\cot x=t}=2y\int_0^\infty \frac{1}{t^2+y^2}\frac{dt}{t^2+1}$$ $$=\frac{2y}{1-y^2}\int_0^\infty \frac{dt}{t^2+y^2}-\frac{2y}{1-y^2}\int_0^\infty \frac{dt}{t^2+1}$$ $$=\frac{2y}{1-y^2}\frac{1}{y} \frac{\pi}{2} -\frac{2y}{1-y^2}\frac{\pi}{2}=\pi\frac{1-y}{1-y^2}=\frac{\pi}{1+y}$$ This gives:$$I(y)=\int \frac{\pi}{1+y} dy =\pi \ln(1+y)+C$$ In order to find the constant we can simply set $y=1$ in the original integral, that is because $\sin^2 x+\cos^2 x$ has a nice value. $$\Rightarrow I(1)=\int_0^\frac{\pi}{2} \ln(1)dx =\pi\ln(1+1)+C \Rightarrow 0=\pi\ln 2 +C \Rightarrow C=-\pi \ln 2$$ $$\Rightarrow I(y)=\pi\ln(1+y) -\pi \ln 2$$

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  • $\begingroup$ $I = \pi \ln(1+y)-\frac {\pi}4 \ln 2?$ $\endgroup$ – C. Cristi Jan 24 '19 at 19:21
  • $\begingroup$ Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration $\endgroup$ – C. Cristi Jan 26 '19 at 8:28
  • $\begingroup$ @C.Cristi I have updated the answer. $\endgroup$ – Zacky Jan 26 '19 at 9:58

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