Question

Intergral log(tanx + cotx) limits0to pie/2

Answer 1

$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\sf{\int \limits^{\frac{\pi}{2}}_{0} \, \log\,(\tan\,x+\cot\,x)\;dx}$

Express tan and cot in terms of sin and cos

$\boxed{ \quad\begin{aligned}\bigstar\quad \tan\:\theta &=\dfrac{\sin\:\theta}{\cos\:\theta}\\\bigstar\quad\cot\:\theta&=\dfrac{cos\:\theta}{\sin\:\theta}\quad\end{aligned}}$

$\mathsf{\int \limits^{\frac{\pi}{2}}_{0} \,\log\,\left(\dfrac{\sin\,x}{\cos\,x}+\dfrac{\cos\,x}{\sin\,x}\right)\;dx}$

$\mathsf{\int \limits^{\frac{\pi}{2}}_{0} \,\log\,\left(\dfrac{\sin^2\,x+\cos^2\,x}{\sin\,x \: \cos\,x}\right)\;dx}$

Using the Trigonometric Identity

$\boxed{\quad\bigstar\quad\sin^2\:\theta + \cos^2\theta =1\quad}$

$\mathsf{\int \limits^{\frac{\pi}{2}}_{0} \,\log\,\left(\dfrac{1}{\sin\,x\:\cos\,x} \right)\;dx}$

$\boxed{\quad\bigstar\quad\sf\log\:\left(\dfrac{a}{b}\right)=\log\:a -\log\:b\quad}$

$\mathsf{-\int \limits^{\frac{\pi}{2}}_{0} \,\log\,(\sin\,x\:\cos\,x)\;dx}$

$\sf{\left[\int\limits^{\frac{\pi}{2}}_{0} \,\log\,\sin\,x\;dx +\int\limits^{\frac{\pi}{2}}_{0}\,\log\,\cos\,x\;dx\right]}$ ———[1]

$\sf{Let\: P = \int \limits^{\frac{\pi}{2}}_{0} \, \log\,\sin\, x\; dx}$

Using the property of Definite Integral

$\boxed{\quad \bigstar\quad\sf\int \limits^{a}_{0}\,f(x)\; dx =\int\limits^{a}_{0}\, f(a-x)\;dx\quad}$

Here,

$f(\cdot)= log\,\sin\,(\cdot)$

$a=\frac{\pi}{2}$

$\implies\sf{\int \limits^{\frac{\pi}{2}}_{0} \, log\,\sin\, x\;dx = \int\limits^{\frac{\pi}{2}}_{0}\,\log\,\sin\,\left(\frac{\pi}{2}-x\right)\;dx}$

$\sf{\sin\,\left(\frac{\pi}{2}-x\right)=\cos\,x}$

$\therefore\;\sf{P = \int\limits^{\frac{\pi}{2}}_{0}\,\log\,\sin\, x\;dx = \int\limits^{\frac{\pi}{2}}_{0}\,\log\,\cos\, x\;dx}$ ———[2]

Substitute in [1]

$\sf{\int \limits^{\frac{\pi}{2}}_{0}\,\log\,(\tan\,x+\cot\,x)\;dx = - (P+P)}$

$\sf{\int \limits^{\frac{\pi}{2}}_{0} \, \log\, (\tan\, x+\cot\, x)\;dx = -2P}$ ———[3]

Do [2]+[2] to find 2P

$\sf{2P =\int \limits^{\frac{\pi}{2}}_{0} \, log\, (\sin\,x +\cos\,x)\; dx}$

Using the Logarithmic Identity

$\boxed{\quad \bigstar \quad \sf \log\: (a+b) = \log\: (ab)\quad }$

$\sf{2P = \int \limits^{\frac{\pi}{2}}_{0} \,\log\,(\sin\,x\:\cos\,x)\; dx}$

$\sf{2P = \int \limits^{\frac{\pi}{2}}_{0} \,log\,\left(\dfrac{2\:\sin\,x\:\cos\,x}{2}\right)\; dx}$

$\boxed{\quad\bigstar\quad\sf\sin\:2x = 2\:\sin\:x\:\cos\: x\quad }$

$\sf{2P = \int \limits^{\frac{\pi}{2}}_{0} \, log\,\left(\dfrac{sin\: 2x}{2} \right)\;dx}$

$\sf{\log\:\left(\dfrac{a}{b}\right) = \log\: (a-b)}$

$\sf{2P = \int\limits^{\frac{\pi}{2}}_{0}\, \log\,(\sin\: 2x - 2)\; dx}$

$\sf{2P = \int\limits^{\frac{\pi}{2}}_{0}\, log\, \sin\, 2x\;-\int\limits^{\frac{\pi}{2}}_{0}\, \log\,2dx}$

$\sf{2P = \int\limits^{\frac{\pi}{2}}_{0}\, log\, \sin\, 2x\; dx-\log\,2\:\int\limits^{\frac{\pi}{2}}_{0}\,1\;dx}$

$\sf{2P =\int\limits^{\frac{\pi}{2}}_{0}\, log\, sin\:2x\;dx -\log\:2\cdot[x\bigm]_{\tiny{0}}^{\tiny{\frac{\pi}{2}}}}$

$\sf{2P = \int\limits^{\frac{\pi}{2}}_{0}\, \log\,\sin\:2x\;dx -\frac{\pi}{2}\,\log\:2}$

$\sf{Let\: Q=\int\limits^{\frac{\pi}{2}}_{0}\, \log\,\sin\:2x\;dx}$

$\sf{(2P = Q -\frac{\pi}{2}\:\log\:2)}$ ———[4]

Do Change of Variable in Q

$\begin{aligned}&t= 2x & dt = 2dx\\x=0;& \quad t=2(0) = 0&dx = \frac{1}{2}\;dt \\x=\dfrac{\pi}{2} ; & \quad t =2(\frac{\pi}{2})= \pi& \end{aligned}$

$\sf{Q=\dfrac{1}{2}\:\int\limits^{\pi}_{0}\, \log\,\sin\,t\; dt}$ ———[5]

$\sf{Q =\cancel{2}\,\left[ \dfrac{1}{\cancel{2}}\:\int\limits^{\pi}_{0}\, log\,\sin\, t\; dt \right]}$

$\sf{Q =\int\limits^{\pi}_{0}\, log\,\sin\,t\; dt}$

Using the Property of Definite Integrals

$\boxed{\begin{minipage}{10cm}\begin{aligned}\quad\bigstar\quad \int\limits^{2a}_{0}\,f(x)\;dx&=\int\limits^{a}_{0}\,f(x)\;dx+\int\limits^{a}_{0}\,f(2a-x)\; dx\\\\\quad \rm \quad \int\limits^{a}_{0}\,f(x)\;dx \; &\begin{cases}=2\:\int\limits^{a}_{0}\, f(x)\;dx&\text{If} f(2a-x)\;dx =f(x)\\\atop\\=0&\text{If} f(2a-x)\;dx =-f(x)\end{cases}\end{aligned}\end{minipage}}$

Here,

$f(\cdot) = \log\,\sin\,(\cdot)$

$2a=\pi\implies a =\frac{\pi}{2}$

$\sf{As\:\log\,\sin\, t =\log\,\sin\,(\frac{\pi}{2}-t)}$ (From [2])

$\therefore \bigstar\quad \sf{ \int\limits^{\pi}_{0}\,\log\,\sin\,t\; dt = 0}$

$\sf{Q = 0 +\int\limits^{\frac{\pi}{2}}_{0}\, log\,sin\,(\frac{\pi}{2}-t)\; dt }$

$\sf{Q = \int\limits^{\frac{\pi}{2}}_{0}\,log\,sin\,(\frac{\pi}{2}-t)\;dt = P}$

Substitute Q in [4]

$\sf{2P = P -\frac{\pi}{2}\:\log\:2}$

$\sf{P =-\frac{\pi}{2}\:\log\:2}$

Substitute in [3]

$\sf{\int \limits^{\frac{\pi}{2}}_{0}\,\log\, (\tan\,x+\cot\,x)\;dx = -2(-\frac{\pi}{2} \:\log\:2)}$

$\sf{\int \limits^{\frac{\pi}{2}}_{0}\,\log\, (\tan\,x+\cot\,x)\; dx =\pi\:\log\:2}$