请教两个求定积分的题

abababa

1.$\int_{0}^{\frac{\pi}{2}}\frac{x\cos x\sin x}{4\cos^2 x+\sin^2 x}dx$
2.$\int_{0}^{\frac{\pi}{2}}\sin^2 x\ln(\sin x)dx$

战巡

回复 1# abababa


\[\int_0^{\frac{\pi}{2}}\frac{x\cos(x)\sin(x)}{3\cos^2(x)+1}dx=\int_0^{\frac{\pi}{2}}dx\int_0^x\frac{\cos(x)\sin(x)}{3\cos^2(x)+1}dy=\int_0^{\frac{\pi}{2}}dy\int_y^{\frac{\pi}{2}}\frac{\cos(x)\sin(x)}{3\cos^2(x)+1}dx\]
\[=\int_0^{\frac{\pi}{2}}\frac{1}{6}[\ln(5+3\cos(2y))-\ln(2)]dy=\frac{1}{6}\int_0^{\frac{\pi}{2}}\ln(5+3\cos(2y))dy-\frac{\pi}{12}\ln(2)\]
其中
\[\int_0^{\frac{\pi}{2}}\ln(5+3\cos(2y))dy=\int_0^{\frac{\pi}{2}}\ln[\frac{2(4+\tan^2(y))}{1+\tan^2(y)}]dy\]
\[=\frac{\pi}{2}\ln(2)+\int_0^{\frac{\pi}{2}}\ln(4+\tan^2(y))dy-\int_0^{\frac{\pi}{2}}\ln(1+\tan^2(y))dy\]
令函数：
\[f(a)=\int_0^{\frac{\pi}{2}}\ln(a+\tan^2(y))dy\]
\[f'(a)=\int_0^{\frac{\pi}{2}}\frac{\partial}{\partial a}\ln(a+\tan^2(y))dy=\int_0^{\frac{\pi}{2}}\frac{dy}{a+\tan^2(y)}\]
\[=\frac{1}{a-1}[y-\frac{1}{\sqrt{a}}\arctan[\frac{\tan(y)}{\sqrt{a}}]]|^{\frac{\pi}{2}}_0=\frac{\pi}{2(a-1)}(1-\frac{1}{\sqrt{a}})\]
由于不难证明：
\[f(0)=\int_0^{\frac{\pi}{2}}2\ln(\tan(y))dy=0\]
因此有：
\[f(a)=\int_0^af'(t)dt=\int_0^a\frac{\pi}{2(t-1)}(1-\frac{1}{\sqrt{t}})dt=\pi\ln(1+\sqrt{a})\]
然后有
\[\int_0^{\frac{\pi}{2}}\ln(5+3\cos(2y))dy=\frac{\pi}{2}\ln(2)+f(4)-f(1)=\frac{\pi}{2}\ln(\frac{9}{2})\]
原式：
\[\int_0^{\frac{\pi}{2}}\frac{x\cos(x)\sin(x)}{3\cos^2(x)+1}dx=\frac{1}{6}[\int_0^{\frac{\pi}{2}}\ln(5+3\cos(2y))dy]-\frac{\pi}{12}\ln(2)=\frac{\pi}{6}\ln(\frac{3}{2})\]


第二题没时间弄了，晚些吧

战巡

回复 1# abababa


\[\int_0^{\frac{\pi}{2}}\sin^2(x)\ln(\sin(x))dx=\int_0^{\frac{\pi}{2}}(1-\cos^2(x))\ln(\sin(x))dx\]
\[=-\frac{\pi}{2}\ln(2)-\int_0^{\frac{\pi}{2}}\cos^2(x)\ln(\sin(x))dx\]
然后有：
\[\int_0^{\frac{\pi}{2}}\cos^2(x)\ln(\sin(x))dx-\int_0^{\frac{\pi}{2}}\sin^2(x)\ln(\sin(x))dx=\int_0^{\frac{\pi}{2}}\cos(2x)\ln(\sin(x))dx\]
\[=[-\frac{1}{2}x-\frac{1}{4}\sin(2x)+\frac{1}{2}\sin(2x)\ln(\sin(2x))]^{\frac{\pi}{2}}_0=-\frac{\pi}{4}\]
于是：
\[\begin{cases}
\int_0^{\frac{\pi}{2}}\sin^2(x)\ln(\sin(x))dx+\int_0^{\frac{\pi}{2}}\cos^2(x)\ln(\sin(x))dx=-\frac{\pi}{2}\ln(2)\\
\int_0^{\frac{\pi}{2}}\sin^2(x)\ln(\sin(x))dx-\int_0^{\frac{\pi}{2}}\cos^2(x)\ln(\sin(x))dx=\frac{\pi}{4}
\end{cases}\]
可得
\begin{cases}
\int_0^{\frac{\pi}{2}}\sin^2(x)\ln(\sin(x))dx=\frac{\pi}{8}-\frac{\pi}{4}\ln(2)\\
\int_0^{\frac{\pi}{2}}\cos^2(x)\ln(\sin(x))dx=-\frac{\pi}{8}-\frac{\pi}{4}\ln(2)
\end{cases}


其实一开始我还想过一种更一般的做法，不过涉及的高等内容比较多
令函数：
\[f(n)=\int_0^{\frac{\pi}{2}}\sin^n(x)dx=\frac{\sqrt{\pi}}{2}\frac{\Gamma(\frac{1}{2}+\frac{n}{2})}{\Gamma(1+\frac{n}{2})}\]
有：
\[f'(n)=\int_0^{\frac{\pi}{2}}\frac{\partial}{\partial n}\sin^n(x)dx=\int_0^{\frac{\pi}{2}}\ln(\sin(x))\sin^n(x)dx\]
另一方面有
\[f'(n)=\frac{d}{dn}\frac{\sqrt{\pi}}{2}\frac{\Gamma(\frac{1}{2}+\frac{n}{2})}{\Gamma(1+\frac{n}{2})}=\frac{\sqrt{\pi}}{2}\frac{\frac{1}{2}\Gamma'(\frac{1}{2}+\frac{n}{2})\Gamma(1+\frac{n}{2})-\frac{1}{2}\Gamma(\frac{1}{2}+\frac{n}{2})\Gamma'(1+\frac{n}{2})}{\Gamma^2(1+\frac{n}{2})}\]
\[=\frac{\sqrt{\pi}}{4}\frac{\Gamma(\frac{1}{2}+\frac{n}{2})}{\Gamma(1+\frac{n}{2})}[\psi(\frac{1}{2}+\frac{n}{2})-\psi(1+\frac{n}{2})]\]
其中$\psi(x)$为双伽马函数，有：
\[\psi(x)=\frac{d}{dx}\ln(\Gamma(x))=\frac{\Gamma'(x)}{\Gamma(x)}\]
而对于$\psi(x)$来说，如果带入整数和半整数，有：
\begin{cases}
\psi(n)=\sum_{k=1}^{n-1}\frac{1}{k}-\gamma\\
\psi(n+\frac{1}{2})=\sum_{k=1}^n\frac{2}{2k-1}-2\ln(2)-\gamma
\end{cases}
其中$\gamma\approx 0.577216...$为欧拉常数

就原题来说，如果带入$n=2$，有：
\[\int_0^{\frac{\pi}{2}}\ln(\sin(x))\sin^2(x)dx=\frac{\sqrt{\pi}}{4}\frac{\Gamma(\frac{1}{2}+1)}{\Gamma(1+1)}[\psi(\frac{1}{2}+1)-\psi(1+1)]\]
\[=\frac{\sqrt{\pi}}{4}\frac{\frac{\sqrt{\pi}}{2}}{1}[2-2\ln(2)-\gamma-(1-\gamma)]=\frac{\pi}{8}(1-2\ln(2))\]

abababa

谢谢，第一题我开始是这么想的，设原积分为$I$，然后
$I=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\frac{2x\sin 2x}{3\cos 2x+5}d2x=\frac{1}{4}\int_{0}^{\pi}\frac{t\sin t}{3\cos t+5}dt=\frac{1}{4}\int_{-\pi}^{0}\frac{t\sin t}{3\cos t+5}dt$
这样$8I=\int_{-\pi}^{\pi}\frac{t\sin t}{3\cos t+5}dt$
然后令$8I'=\int_{-\pi}^{\pi}\frac{t\cos t}{3\cos t+5}dt$，令$z=e^{it}$
这样$8I'+i8I=\int_{|z|=1}\frac{-iz\ln z}{3\frac{z+1/z}{2}+5}\frac{dz}{iz}=-\int_{|z|=1}\frac{2z\ln zdz}{(3z+1)(z+3)}$
到这里就算不下去了，就是$\ln z$的部分不会算
2楼给的计算看起来也挺复杂的