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[组合] 组合恒等式

\[\sum_{i=1}^{n}\frac{\pmatrix{n\\i}}{i}\]
\[\sum_{i=1}^{n}(-1)^i\frac{\pmatrix{n\\i}}{pn+si}\]

搜狗截图20140308191739.png
2014-3-8 19:18
分享到: QQ空间QQ空间 腾讯微博腾讯微博 腾讯朋友腾讯朋友

本帖最后由 战巡 于 2014-3-9 02:19 编辑

回复 1# 青青子衿


令函数$f(x)=x^{pn-1}(1-x^s)^n$
易证:
\[f(x)=\sum_{k=0}^n (-1)^k C_n^k x^{pn+sk-1}\]
\[\int_0^1f(x)dx=\int_0^1 x^{pn-1}(1-x^s)^ndx=\sum_{k=0}^n\int_0^1(-1)^k C_n^k x^{pn+sk-1}dx=\sum_{k=0}^n(-1)^k \frac{C_n^k}{pn+sk}\]
换元$x^s=t$,得到
\[\int_0^1 x^{pn-1}(1-x^s)^ndx=\frac{1}{s}\int_0^1 t^{\frac{pn}{s}-1}(1-t)^ndt=\frac{1}{s} B(\frac{pn}{s},n+1)=\frac{\Gamma(\frac{pn}{s})\Gamma(n+1)}{s\Gamma(\frac{pn}{s}+n+1)}\]

剩下自己算吧

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回复 2# 战巡
哦,谢谢!

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25228162852.png
2020-9-28 16:29

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本帖最后由 青青子衿 于 2021-6-18 13:21 编辑

1979年国防科技大学数学分析
245240194136.png
2020-10-20 19:42



\begin{align*}
\color{black}{
\begin{split}
\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\frac{\ln^2\left(1-x\right)}{\sqrt{1-x\sin^2y}}\mathrm{d}x\mathrm{d}y
&=48-\frac{4}{3}\pi^{2}-64\ln2+32\ln^2\,\!2\\
&=\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\frac{\ln^2\left(1-u^2\right)}{\sqrt{1-u^2\sin^2y}}\mathrm{d}(u^2)\mathrm{d}y\\
&=2\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\frac{u\ln^2\left(1-u^2\right)}{\sqrt{1-u^2\sin^2y}}\mathrm{d}u\mathrm{d}y\\
&=2\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\frac{\sin\alpha\ln^2\left(1-\sin^2\alpha\right)}{\sqrt{1-\sin^2\alpha\sin^2y}}\mathrm{d}(\sin\alpha)\mathrm{d}y\\
&=8\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\frac{\sin\alpha\cos\alpha\ln^2\left(\cos\alpha\right)}{\sqrt{1-\sin^2\alpha\sin^2\beta}}\mathrm{d}\alpha\mathrm{d}\beta\\
&=8\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\sin\theta\cos\varphi\ln^2\left(\sin\theta\cos\varphi\right)\mathrm{d}\theta\mathrm{d}\varphi\\
&=8\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\sin\theta\cos\varphi\left(\ln\sin\theta+\ln\cos\varphi\right)^2\mathrm{d}\theta\mathrm{d}\varphi\\
&=8\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\sin\theta\cos\varphi\left(\ln^2\sin\theta+\ln^2\cos\varphi+2\ln\sin\theta\ln\cos\varphi\right)\mathrm{d}\theta\mathrm{d}\varphi\\
&=16\int_{0}^{\frac{\pi}{2}}\sin\theta\ln^2\left(\sin\theta\right)dx+16\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\sin\theta\cos\varphi\ln\left(\sin\theta\right)\ln\left(\cos\varphi\right)\mathrm{d}\theta\mathrm{d}\varphi\\
&=16\int_{0}^{\frac{\pi}{2}}\sin\theta\ln^2\left(\sin\theta\right)dx+16\left[\int_{0}^{\frac{\pi}{2}}\sin\theta\ln\left(\sin\theta\right)\mathrm{d}\theta\right]^2\\
\end{split}
}
\end{align*}

\begin{align*}
\color{black}{
\boldsymbol{Qn.39}\qquad\int_0^{+\infty}\arctan\left(\dfrac{2ax}{x^2+b^2}\right)\sin(kx)\mathrm{d}x=\dfrac{\pi}{k}\exp\left(-k\sqrt{a^2+b^2}\right)\sinh(ak)
}\\

\end{align*}

\begin{align*}
\color{black}{
\boldsymbol{Qn.44}\qquad
\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\int_{0}^{1}\int_{0}^{1}r\rho\sqrt{r^{2}-2r\rho\cos\left(\theta-\varphi\right)+\rho^{2}}\,\mathrm{d}r\mathrm{d}{\rho}\mathrm{d}{\theta}\mathrm{d}{\varphi}=\dfrac{128\pi}{45}
}
\end{align*}

\begin{align*}
\left\{
\begin{split}
p&=x+\xi\\
q\,&=y+\eta\\

\psi&=x-\xi\\
\varphi&=y-\eta
\end{split}
\right. \quad
\quad&\Rightarrow\quad
\left\{
\begin{split}
x&=\dfrac{p+\psi}{2}\\
y&=\dfrac{q+\varphi}{2}\\

\xi&=\dfrac{p-\psi}{2}\\
\eta&=\dfrac{q-\varphi}{2}
\end{split}
\right. \\
\\
\left\{
\begin{split}
0&\leqslant\,\!x\leqslant1-y\\
0&\leqslant\,\!y\leqslant1\\

0&\leqslant\xi\leqslant1-\eta\\
0&\leqslant\eta\leqslant1\\
\end{split}
\right.
\quad&\Rightarrow\quad
\left\{
\begin{split}
?&\leqslant\,\!p\leqslant\,?\\
?&\leqslant\,\!q\leqslant\,?\\
?&\leqslant\psi\leqslant\,?\\
?&\leqslant\varphi\leqslant\,?
\end{split}
\right. \\
\end{align*}

\begin{align*}
\color{black}{

\begin{split}
&\qquad\qquad\qquad\qquad\int_{-\infty}^{+\infty}\frac{e^{-\lambda\,\!x}}{\cosh(x+a)\cosh(x+b)\cosh(x+c)}\mathrm{d}x\\
\\
&=\frac{\pi}{\cos\left(\frac{\lambda\pi}{2}\right)}\left[\frac{\exp\left(\lambda a\right)}{\sinh\left(a-b\right)\sinh\left(c-a\right)}+\frac{\exp\left(\lambda b\right)}{\sinh\left(b-a\right)\sinh\left(c-b\right)}+\frac{\exp\left(\lambda c\right)}{\sinh\left(c-a\right)\sinh\left(b-c\right)}\right]
\end{split}}
\end{align*}

\begin{equation*}
\int_{-\infty}^{\infty} \int_{-\infty}^{s_{1}} \cdots \int_{-\infty}^{s_{2n-1}}
\prod_{j=1}^{n}
\cos(s_{2j-1}^{2} - s_{2j}^{2})
\, ds_{2n} \cdots ds_{1} = \frac{\pi^{n}}{2^{2n-1} n! }.
\end{equation*}
\begin{align*}
\color{black}{
\boldsymbol{Qn.45} \qquad
\int_{0}^{\frac{\pi}{4}}\!\!\!\int_{0}^{\frac{\pi}{4}}\!\!\!\int_{0}^{\frac{\pi}{4}}\!\!\!\int_{0}^{1}\!\!\!\int_{0}^{1}\!\!\!\int_{0}^{1}\dfrac{\mathrm{d}u\mathrm{d}v\mathrm{d}w\mathrm{d}x\mathrm{d}y\mathrm{d}z}{(1+u^2+v^2+w^2+\sec^2x+\sec^2y+\sec^2z)^2}=\dfrac{\pi^5}{20480}
}
\end{align*}

$$\begin{align*}
&\qquad\begin{split}
\boldsymbol{B}&=\{b_{i,j}\}\quad(1\leqslant\,\!i\leqslant\,\!j\leqslant\,\!n)\\
\Downarrow\,&\\
\boldsymbol{A}&=\{a_{i,j}\}\quad(1\leqslant\,\!i\leqslant\,\!j\leqslant\,\!n)
\end{split}
\\
\\
&\left\{
\begin{split}
a_{i,i}&=1\qquad\quad(1\leqslant\,\!i\leqslant\,\!n)\\
a_{i,i+1}&=-\frac{b_{i,i+1}}{b_{i+1,i+1}}\qquad(1\leqslant\,\!i\leqslant\,\!n-1)\\
\\
a_{i,j}\,\,&=-\frac{b_{i,j}+\sum\limits_{i<k<j}a_{i,k}b_{k,j}}{b_{j,j}}\quad(1\leqslant\,\!i\leqslant\,\!j-2\leqslant\,\!n-2)\\
\\
a_{i,j}&=0\qquad(i>j)
\end{split}
\right.
\end{align*}$$


\[
\color{black}{
\begin{gather*}
\boldsymbol{Qn.47}\qquad\\
\\
\int_0^{+\infty}\dfrac{\exp\left(\cos\,\!ax\right)\cos\left[\sin(ax)+bx\right]}{x^4+c^4}\mathrm{d}x\qquad(a>0,b>0,c>0)\\
\\
=\dfrac{\pi}{2\sqrt{2}\,c^3}\exp\Big[\exp(-A)\cos(A)-B\Big]\Big\{\cos\Big[\exp(-A)\sin(A)+B\Big]+\sin\Big[\exp(-A)\sin(A)+B\Big]\Big\}\\
\\
\left(A=\dfrac{\,\,ac}{\sqrt{2}};\,B=\dfrac{\,\,bc}{\sqrt{2}}\,\right)
\end{gather*}
}
\]


\begin{align*}   
\color{black}{
\begin{split}   
\boldsymbol{Qn.52}&\\
\boldsymbol{K}(k)&=\int_0^1\dfrac{\mathrm{d}t}{\sqrt{(1-t^2)(1-k^2t^2)}}\\   
     
\vartheta_3(q)&=\sum_{n=-\infty}^{+\infty}q^{n^2}\\   
\\   
I&=\int_0^1\dfrac{[x(1-x)]^{-11/12}}{\left(\vartheta_3\left\{\exp\left[-\frac{\pi\,\boldsymbol{K}\big(\sqrt{1-x}\big)}{\boldsymbol{K}\big(\sqrt{x}\big)}\right]\right\}\right)^3}\mathrm{d}x=\dfrac{2\sqrt[3]{2}}{\sqrt{3}}\pi^2   
\end{split}}
\end{align*}


\begin{align*}   
\color{black}{  
\begin{split}   
\boldsymbol{Qn.81}\qquad\qquad\qquad\qquad&\\
\\
\int_{0}^{+\infty}\frac{\cos\left(a+b\ln\,\!x\right)}{t^{2}+x^{2}}\mathrm{d}x&=\frac{\pi\cos\left(a+b\ln t\right)}{2t\cosh\left(\frac{b\pi}{2}\right)}   
\end{split}}  
\end{align*}

\begin{align*}  
\color{black}{  
\begin{split}   
\boldsymbol{Qn.61}&\\  
&
\int_{0}^{+\infty}\!\!\int_{0}^{+\infty}\!\!\int_{0}^{+\infty}\!\frac{\left(x^{2}+y^{2}+z^{2}\right)^{\frac{m}{2}}\,}{\left(x^{2}+y^{2}+z^{2}+a^{2}\right)^{\frac{m+5}{2}}\,\,}\,\,\mathrm{d}x\mathrm{d}y\mathrm{d}z=\dfrac{\pi}{2(m+3)a^2}

\end{split}}  
\end{align*}

\begin{align*}
\int _{-\infty }^{+\infty }\int _{-\infty }^{+\infty }e^{-\gamma \sqrt{x^2+y^2\,\,}}\!\cos (\alpha\,\!x) \cos (\beta\,\!y)\,\mathrm{d}x\mathrm{d}y=\frac{2 \pi\gamma}{\left(\alpha^2+\beta^2+\gamma^2\right)^{3/2}}
\end{align*}

\begin{align*}   
\color{black}{   
\begin{split}     
\boldsymbol{Qn.65}&\\   
&
\int_{0}^{+\infty}\!\frac{xJ_0(ux)}{\,\,\left(x^2+v^{2}\right)^{5/2}}\,\mathrm{d}x=\dfrac{1+uv}{3v^3\mathrm{e}^{uv}}

\end{split}}   
\end{align*}

\begin{align*}   
\color{black}{   
\begin{split}     
\boldsymbol{Qn.66}&\\   
&\int_{-\infty}^{+\infty}\!\int_{-\infty}^{+\infty}\frac{1}{\sqrt{\left(x^{2}+y^{2}+a^{2}\right)^{5}\left[\left(x-u\right)^{2}+\left(y-v\right)^{2}+b^{2}\right]}}\mathrm{d}x\mathrm{d}y\\
&=\frac{2\pi}{3a^{3}}\left\{\frac{1}{\sqrt{\left(a+b\right)^{2}+u^{2}+v^{2}}}+\frac{a\left(a+b\right)}{\sqrt{\left[\left(a+b\right)^{2}+u^{2}+v^{2}\right]^{3}}}\right\}
\end{split}}   
\end{align*}

\begin{align*}
\color{black}{
yy''+ (y')^2=(y'\ln\,\!y)^3\\
\mathscr{F}}
\end{align*}

\begin{align*}   
\color{black}{   
\begin{split}      
\boldsymbol{Qn.66}&\\   
&  
\int_{-\infty}^{+\infty}\dfrac{\cos\left(\frac{x^2}{\pi}\right)}{\cosh(x+a)\cosh(x+b)\cosh(x+c)}\mathrm{d}x
  
\end{split}}   
\end{align*}




\begin{align*}   
f(x,y)&=\dfrac{1}{\sqrt{\left(x^{2}+y^{2}+a^{2}\right)^{5}}}\qquad\qquad\,g(x,y)=\dfrac{1}{\sqrt{x^{2}+y^{2}+a^{2}}}\\
\hat{f}(\xi,\eta)&=\dfrac{1+a\sqrt{\xi^2+\eta^2}}{3a^3\exp\left(\,a\sqrt{\xi^2+\eta^2}\,\right)}\quad\,\,\quad\,g(\xi,\eta)=\dfrac{1}{\sqrt{\xi^2+\eta^2}\exp\left(\,b\sqrt{\xi^2+\eta^2}\,\right)}\\

\varphi(\xi,\eta)&=\dfrac{\exp\left[\,-(a+b)\sqrt{\xi^2+\eta^2}\,\right]}{\sqrt{\xi^2+\eta^2}}\quad\quad
\psi(\xi,\eta)=\exp\left[\,-(a+b)\sqrt{\xi^2+\eta^2}\,\right]\\
\hat{f}\cdot\hat{g}&=\dfrac{\exp\left[\,-(a+b)\sqrt{\xi^2+\eta^2}\,\right]}{3a^3}\left(\dfrac{1}{\sqrt{\xi^2+\eta^2}}+a\right)\\
&=\dfrac{1}{3a^3}\Big[\varphi(\xi,\eta)+a\cdot\psi(\xi,\eta)\Big]\\
\\
&\qquad\qquad\qquad\qquad\,\,\,f*g=\mathscr{F}^{-1}(\hat{f}\cdot\hat{g})\\

\\

I&=\int_{-\infty}^{+\infty}\!\int_{-\infty}^{+\infty}\frac{1}{\sqrt{\left(x^{2}+y^{2}+a^{2}\right)^{5}\left[\left(x-u\right)^{2}+\left(y-v\right)^{2}+b^{2}\right]}}\mathrm{d}x\mathrm{d}y\\
&=\int_{-\infty}^{+\infty}\!\int_{-\infty}^{+\infty}f\big(x,y\big)g\big(x-u,y-v\big)\mathrm{d}x\mathrm{d}y\\




&=\int_{-\infty}^{+\infty}\!\int_{-\infty}^{+\infty}\hat{f}\big(\xi,\eta\big)\hat{g}\big(\xi,\eta\big)\,\mathrm{e}^{-2\pi\,\!i(u\xi+v\eta)}\mathrm{d}\xi\mathrm{d}\eta\\
&=\dfrac{1}{3a^3}\int_{-\infty}^{+\infty}\!\int_{-\infty}^{+\infty}\varphi(\xi,\eta)\,\mathrm{e}^{-2\pi\,\!i(u\xi+v\eta)}\mathrm{d}\xi\mathrm{d}\eta\\
&\qquad\qquad\qquad+\dfrac{1}{3a^2}\int_{-\infty}^{+\infty}\!\int_{-\infty}^{+\infty}\psi(\xi,\eta)\,\mathrm{e}^{-2\pi\,\!i(u\xi+v\eta)}\mathrm{d}\xi\mathrm{d}\eta\\
&=\frac{1}{3a^{3}}\cdot\frac{2\pi}{\sqrt{\left(a+b\right)^{2}+u^{2}+v^{2}}}+\frac{1}{3a^{2}}\cdot\frac{2\pi(a+b)}{\sqrt{\left[\left(a+b\right)^{2}+u^{2}+v^{2}\right]^{3}}}\\
&=\frac{2\pi}{3a^{3}}\left\{\frac{1}{\sqrt{\left(a+b\right)^{2}+u^{2}+v^{2}}}+\frac{a\left(a+b\right)}{\sqrt{\left[\left(a+b\right)^{2}+u^{2}+v^{2}\right]^{3}}}\right\}

\end{align*}


\begin{align*}
\color{black}{
\boldsymbol{Qn.76}\qquad
\int_0^{+\infty} \frac{x\sin(a\tan x)}{x^2+b^2}\mathrm{d}x =\frac{\pi}{2}\left(e^{-a\tanh b}-e^{-a}\right) \qquad a,b>0
}
\end{align*}




\begin{align*}     
\color{black}{   
\begin{split}     
\boldsymbol{Qn.82}\qquad\qquad\qquad\qquad\qquad\qquad&\\  
\\  
\int_{0}^{+\infty}\arctan\left(\frac{1}{1+x^{2}}\right)\cos\left(x\right)\mathrm{d}x&=
\pi\exp\left(-\sqrt[4]{2}\cos\frac{\pi}{8}\right)\sin\left(\sqrt[4]{2}\sin\frac{\pi}{8}\right)
  
\end{split}}   
\end{align*}

\begin{align*}      
\color{black}{   
\begin{split}      
\boldsymbol{Qn.84}\qquad\qquad\qquad\qquad\qquad\qquad&\\   
\\   
\int_{\!-\pi}^{\pi}\dfrac{\mathrm{e}^{\sin(x)+\cos(x)}\cos(\sin\,\!x)}{\mathrm{e}^x+\mathrm{e}^{\sin\,\!x}}\mathrm{d}x&=
\pi
   
\end{split}}   
\end{align*}


\begin{align*}      
\color{black}{     
\begin{split}      
\boldsymbol{Qn.86}\qquad\qquad\qquad\qquad&\\   
\\   
\int_0^{\!\frac\pi2}\frac{\sin(x)\cos(\tan x)}{(1+\cos^2x)e^{\sin x}}\,\mathrm{d}x&=\frac{\pi}{4\,e^{2\sqrt2}}
   
\end{split}}     
\end{align*}





\begin{gather*}
\displaystyle \int_0^{2\pi}e^{\cosh(\cos x)\cos(\sin x)} \cos(\sinh(\cos x)\sin(\sin x)) \ dx = 2 \pi e\\
\displaystyle  \int^{\infty}_{0} \dfrac{x^2+3x+3}{(x+1)^3}e^{-x}\sin x \; \mathrm{d}x = \dfrac{1}{2}
\end{gather*}


\begin{align*}

\end{align*}

\begin{align*}      
\color{black}{   
\begin{split}      
&\qquad\qquad\boldsymbol{Qn.92}\\   
\\   
&\qquad\qquad\qquad\qquad\qquad\int_{0}^{+\infty}\frac{x^{\alpha}\cosh\left(\gamma+\beta\ln x\right)}{t^{2}+x^{2}}\,\mathrm{d}x\\
\\
&\frac{\pi t^{\alpha-1}\left[\cos\Big(\frac{\pi\alpha}{2}\Big)\cos\Big(\frac{\pi\beta}{2}\Big)\cosh\left(\gamma+\beta\ln t\right)+
\sin\Big(\frac{\pi\alpha}{2}\Big)\sin\Big(\frac{\pi\beta}{2}\Big)\sinh\left(\gamma+\beta\ln t\right)\right]}{\cos\left(\pi\alpha\right)+\cos\left(\pi\beta\right)}
   
\end{split}}   
\end{align*}


\begin{align*}      
\color{black}{   
\begin{split}      
&\quad\boldsymbol{Qn.93}\\   
\\   
&\,\,\int_{0}^{+\infty}\frac{\cos\left(3x\right)\cos\left(2\sqrt{x^2+5}\right)}{\left(x^{2}+1\right)\left(x^{2}+5\right)}\,\mathrm{d}x\\
\\
&=-\frac{\pi}{40}\left[\frac{\sqrt{5}}{\exp\left(3\sqrt{5}\right)}-\frac{5\cos(4)}{\exp\left(3\right)}\right]
   
\end{split}}   
\end{align*}

\begin{align*}      
\color{black}{     
\begin{split}      
&\boldsymbol{Qn.94}\\   
\\   
&\int_0^{2\pi}e^{\cosh(\cos x)\cos(\sin x)} \cos(\sinh(\cos x)\sin(\sin x))\,\mathrm{d}x = 2 \pi e
   
\end{split}}     
\end{align*}




\begin{align*}
\color{black}{
\begin{split}
I&=\int_{0}^{1}\boldsymbol{K}[x]\ln\left(x\right)\,\mathrm{d}x\\
&=\int_{0}^{1}\boldsymbol{K}(\sqrt{x})\ln\left(x\right)\,\mathrm{d}x\\
&=\int_{0}^{1}\ln\left(x\right)\mathrm{d}x\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{1-x\sin^2y}}\,\mathrm{d}y\\
&=\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\frac{\ln\left(x\right)}{\sqrt{1-x\sin^2y}}\,\mathrm{d}x\mathrm{d}y\\
&=\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\frac{\ln\left(u^2\right)}{\sqrt{1-u^2\sin^2y}}\mathrm{d}(u^2)\mathrm{d}y\\
&=4\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\frac{u\ln\left(u\right)}{\sqrt{1-u^2\sin^2y}}\mathrm{d}u\mathrm{d}y\\
&=4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\frac{\sin\alpha\ln\left(\sin\alpha\right)}{\sqrt{1-\sin^2\alpha\sin^2y}}\mathrm{d}(\sin\alpha)\mathrm{d}y\\
&=4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\frac{\sin\alpha\cos\alpha\ln\left(\sin\alpha\right)}{\sqrt{1-\sin^2\alpha\sin^2\beta}}\mathrm{d}\alpha\mathrm{d}\beta\\
&=2\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\cos\alpha\ln\left(1-\cos^2\alpha\right)\cdot\frac{\sin\alpha}{\sqrt{1-\sin^2\alpha\sin^2\beta}}\mathrm{d}\alpha\mathrm{d}\beta\\

&=2\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\sin\theta\cos\varphi\ln\left(1-\sin^2\theta\cos^2\varphi\right)\mathrm{d}\theta\mathrm{d}\varphi\\
&=4\int_{0}^{\frac{\pi}{2}}\Big[\sin(\varphi)\arctan(\cot\varphi)-\cos (\varphi )\Big]\mathrm{d}\varphi\\
&=4\int_{0}^{\frac{\pi}{2}}\Big[(\tfrac{\pi}{2}-\varphi)\sin(\varphi)-\cos (\varphi )\Big]\mathrm{d}\varphi\\
&=4\cdot\dfrac{\pi-4}{2}=2\pi-8
\end{split}
}
\end{align*}


\begin{align*}        
\color{black}{      
\begin{split}        
&\boldsymbol{Qn.96}\\     
\\     
&\quad\,\,\int_0^{+\infty}\!\int_0^1\int_0^1\exp(-x^2z^2) \sin(y^2z^2)\,\mathrm{d}x\mathrm{d}y\mathrm{d}z\\
\\
&=
\frac{\sqrt{2\pi}}{8}\arcsin \left( \sqrt{2}-1 \right)+\frac{\sqrt{\pi}}{2}\arctan \left( \sqrt{\sqrt{2}-1} \,\right)\\
&\qquad\quad\,\,
+\frac{\sqrt{2\pi}}{8}\operatorname{artanh}\left[ \sqrt{2\left( \sqrt{2}-1 \right)}\,\right]-\frac{\pi \sqrt{2\pi}}{16}

\end{split}}      
\end{align*}


\begin{align*}         
\color{black}{      
\begin{split}         
&\boldsymbol{Qn.97}\\      
\\      
&\int_{-1}^{1}\frac{\exp\left[\exp(x)\cos\left(\sqrt{1-x^{2}}\right)\right]\sin\left[\exp(x)\sin\left(\sqrt{1-x^{2}}\right)\right]}{t^{2}-2tx+1}\,\mathrm{d}x
\\
\\
&\qquad\qquad\qquad\qquad\,\,\,=
\frac{\pi\left[\exp\left(\exp{t}\right)-e\right]}{2t}

\end{split}}      
\end{align*}


\begin{align*}

\end{align*}

\begin{align*}         
\color{black}{        
\begin{split}         
&\quad\boldsymbol{Qn.100}\\      
\\      
&\int_0^{\pi/2}\frac{\sin\frac x2}{\sqrt{\cos x}}\arctan\left(\frac{2\sin x}{2\cos x-1}\right)\,\mathrm{d}x\\
\\
&\qquad\qquad=\sqrt{2}\pi\ln\left(\dfrac{1+\sqrt5}{2}\right)-\frac{\pi}{\sqrt{2}}\ln\Big(2+\sqrt{3}\Big)
  
\end{split}}        
\end{align*}


\begin{align*}
\left\{
\begin{split}
&\qquad\dfrac{\pi\,\!b}{24}(9 a^2 - b^2)&&b\leqslant\,\!a\\
&\dfrac{\pi}{48}\Big[24 a^3 - (3 a - b)^3\Big]&\quad&a\leqslant\,\!b\leqslant3a\\
&\qquad\qquad\dfrac{\pi\,\!a^3}{2}&&3a\leqslant\,\!b\\
\end{split}\right.
\end{align*}


\begin{align*}
\color{black}{\int_0^{+\infty}\dfrac{\sin^2(\tan\,\!x)}{x^2}\mathrm{d}x\,\overset{?}{=}\,\dfrac{\pi}{2}}
\end{align*}




\begin{align*}         
\color{black}{        
\begin{split}         
&\mathrm{P.V.}\int_{-\infty}^{+\infty}\!\dfrac{\cos\left(\frac{x^2}{\pi\,}\right)}{\sinh(x+a)}\mathrm{d}x\qquad\,\,\\
\\
&\color{red}{\quad=\dfrac{\sin\left(\frac{a^2}{\pi}\right)\cosh(a)}{\sinh(a)}\,\pi\,}
\end{split}}        
\end{align*}



\begin{align*}         
\color{black}{        
\begin{split}   
\int_0^{+\infty}\frac{e^{-a x^2(x^2-\pi^2)}\cos(2\pi a x^3)}{\cosh x}dx=\frac{\pi}{2}e^{-\pi^4 a/16}
\end{split}}        
\end{align*}


\begin{align*}         
\color{black}{        
\begin{split}   
\lim\limits_{n\to+\infty}n\left\{\int_{0}^{1}f\left(x\right)\mathrm{d}x-\sum_{k=1}^{n}\left[\frac{k^{m}-\left(k-1\right)^{m}}{n^{m}}\right]f\left[\left(\frac{k-1}{n}\right)^{m}\right]\right\}
\end{split}}        
\end{align*}

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