Integral value of an integral

$$ Let $t=\sec x\;\rightarrow\;dt=\tan x\sec x\,dx$ and for $0<x<\frac{\pi}{2}$ is corresponding to $1<t<\infty$ , then $$ $\begin{aligned} \int_0^{\frac{\pi}{2}}\tan x\sqrt{\cos x}\ln(\sec x)\,dx&=\int_0^{\frac{\pi}{2}}\frac{\tan x\ln(\sec x)}{\sqrt{\sec x}}\,dx\\ &=\int_1^{\infty}t^{-\frac{3}{2}}\ln t\,dt. \end{aligned}$ $$ The last form integral can be solven by using IBP. Let $u=\ln t\;\rightarrow\;du=\frac{1}{t}\,dt$ and $dv=t^{-\frac{3}{2}}\,dt\;\rightarrow\;v=-2t^{-\frac{1}{2}}$ . Therefore $$ $\begin{aligned} \int_1^{\infty}t^{-\frac{3}{2}}\ln t\,dt&=-2\lim_{a\to\infty}\left.t^{-\frac{1}{2}}\ln t\right|_{t=1}^a+2\int_1^\infty t^{-\frac{1}{2}\cdot}\frac{1}{t}\,dt\\ &=0+2\int_1^\infty t^{-\frac{3}{2}}\,dt\\ &=-4\lim_{a\to\infty}\left.t^{-\frac{1}{2}}\right|_{t=1}^a\\ &=-4(0-1)\\ &=\boxed{4} \end{aligned}$ $$ $\text{\# }\mathbb{Q.E.D.}\text{ \#}$

Let $lnsecx=z$ .

Then $z=\lim_{x\to0^{+}}lnsecx=0$

$z=\lim_{x\to\infty^-}lnsecx=\infty$ .

$\int^\frac{\pi}{2}_0 tanx\,\sqrt{cosx}\,lnsecx\,dx = \int^\infty_0 z\,e^\frac{-z}{2}\,dz=\textbf{I}$ $(say)$

$\textbf{I}=\lim_{z\to\infty^-}-2z\,e^\frac{-z}{2}+\lim_{z\to0^+}2z\,e^\frac{-z}{2}+2\int^\infty_0 e^\frac{-z}{2} dz$

$\rightarrow\textbf{I}=\boxed{4}$

The integral is $I=-\dfrac{1}{4}\int_0^{\pi/2} 2\sin(t)\cos^{-1/2}(t)\ln(\cos^2(t))\text{dt}$ COnsider the beta function $B(1,x)=\int_0^{\pi/2} 2\sin(t)\cos^{2x-1}(t)\text{dt}$ d.w.r.x $B'(1,x)=\int_0^{\pi/2}2\sin(t)\cos^{2x-1}(t)\ln(\cos^2(t))\text{dt}$ Now $B'(1,x)=\dfrac{d}{dx} \dfrac{\Gamma(1)\Gamma(x)}{\Gamma(x+1)}=\dfrac{d}{dx} \dfrac{1}{x}=\dfrac{-1}{x^2}$ putting x=1/4 $I=-\dfrac{B'(1,1/4)}{4}=\dfrac{4^2}{4}=\boxed{4}$