int_0^{pi / 2} log |tan x+cot x| , dx= | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int_0^{\pi / 2} \log |	an x + \cot x| \, dx$. Since $	an x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x +

Vedclass · Accepted Answer

$\pi \log 2$

Vedclass · Answer

(A) Let $I = \int_0^{\pi / 2} \log |	an x + \cot x| \, dx$. Since $	an x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$. Thus,$I = \int_0^{\pi / 2} \log |\frac{2}{\sin 2x}| \, dx = \int_0^{\pi / 2} (\log 2 - \log |\sin 2x|) \, dx$. $I = \int_0^{\pi / 2} \log 2 \, dx - \int_0^{\pi / 2} \log |\sin 2x| \, dx$. $I = \frac{\pi}{2} \log 2 - \int_0^{\pi / 2} \log |\sin 2x| \, dx$. Let $2x = t$,then $2 \, dx = dt$,so $dx = \frac{dt}{2}$. When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\pi$. $\int_0^{\pi / 2} \log |\sin 2x| \, dx = \frac{1}{2} \int_0^{\pi} \log |\sin t| \, dt = \frac{1}{2} 	imes 2 \int_0^{\pi / 2} \log |\sin t| \, dt = \int_0^{\pi / 2} \log |\sin t| \, dt$. Using the property $\int_0^{\pi / 2} \log |\sin t| \, dt = -\frac{\pi}{2} \log 2$. Therefore,$I = \frac{\pi}{2} \log 2 - (-\frac{\pi}{2} \log 2) = \frac{\pi}{2} \log 2 + \frac{\pi}{2} \log 2 = \pi \log 2$.

$\int_0^{\pi / 2} \log |\tan x+\cot x| \, dx=$

Similar Questions

$\int_{-1}^{1} \log \left( \frac{1+x}{1-x} \right) \, dx = $

The value of the integral $\int \limits_{-\pi / 2}^{\pi / 2} \frac{d x}{\left(1+e^{x}\right)\left(\sin ^{6} x+\cos ^{6} x\right)}$ is equal to

If $f$ is a continuous function and $f(x+T)=f(x)$ for all $x \in R$,it is given that $\int_0^{NT} f(t) dt = N \int_0^T f(t) dt$ (where $N$ is a natural number). Then,evaluate $\int_0^{50\pi} \sqrt{1-\cos 2x} dx$.

Let $f(x) = \frac{\sin x}{x}$,then $\int_{0}^{\frac{\pi}{2}} f(x) f\left(\frac{\pi}{2} - x\right) dx =$

$\int_{-2}^0 (x^3+3x^2+3x+3+(x+1) \cos(x+1)) \, dx$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module