# 请问这个积分问题怎么求，谢谢

\displaystyle\int_0^{\frac{\pi}{2}}\ln (a^2sin^2x+b^2cos^2x)dx \\(a^2+b^2\neq 0)

\displaystyle \mathrm{if}\ \ |a|=0，|b|>0，b^2=|b^2|，\\{}\\ \int_0^{\frac{\pi}{2}}\ln(b^2\cos^2x)dx=\pi\ln|b|+2\int_0^{\frac{\pi}{2}}\ln(\cos x)dx\\{}\\ =\pi\ln|b|-\pi\ln2\\{}\\ =\pi\ln\frac{|b|}{2}.\\{}\\ --------------------\\{}\\ \mathrm{if}\ \ |b|=0，|a|>0，\\{}\\ \int_0^{\frac{\pi}{2}}\ln(a^2\sin^2x)dx=\pi\ln\frac{|a|}{2}.\\{}\\ --------------------\\{}\\ \mathrm{if}\ \ |a|>0，|b|>0，\mathrm{set}\ \ I(b)=\int_0^{\frac{\pi}{2}}\ln(|a|^2\sin^2x+|b|^2\cos^2x)dx\\{}\\ \Rightarrow I'(b)=\int_0^{\frac{\pi}{2}}\frac{2|b|\cos^2x}{a^2\sin^2x+b^2\cos^2x}dx\\{}\\ =\frac{2}{|b|}\int_0^\frac{\pi}{2}\frac{1}{1+\left(|\frac{a}{b}|\tan x\right)^2}dx\\{}\\ =\frac{\pi}{|a|+|b|}，\\{}\\ \mathrm{and}\ \because I(0)=\int_0^\frac{\pi}{2}\ln(a^2\sin^2x)dx=\pi\ln\frac{|a|}{2}，\\{}\\ \therefore I(b)=\int_0^{|b|}\frac{\pi dt}{|a|+t}+\pi\ln\frac{|a|}{2}=\pi\ln\frac{|a|+|b|}{2}，\\{}\\ \Rightarrow \int_0^\frac{\pi}{2}\ln(a^2\sin^2x+b^2\cos^2x)dx=\pi\ln\frac{|a|+|b|}{2}.

1个月前 评论