int_{0}^{pi} [cot x] dx = | vedclass

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

Question

(C) Let $I = \int_{0}^{\pi} [\cot x] dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get: $I = \int_{0}^{\pi} [\cot(\pi - x

Vedclass · Accepted Answer

$-\frac{\pi}{2}$

Vedclass · Answer

(C) Let $I = \int_{0}^{\pi} [\cot x] dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get: $I = \int_{0}^{\pi} [\cot(\pi - x)] dx = \int_{0}^{\pi} [-\cot x] dx$. Adding the two expressions for $I$: $2I = \int_{0}^{\pi} ([\cot x] + [-\cot x]) dx$. We know that for any $y 
otin \mathbb{Z}$,$[y] + [-y] = -1$,and for $y \in \mathbb{Z}$,$[y] + [-y] = 0$. The function $\cot x$ is not defined at $x=0, \pi/2, \pi$. At $x=\pi/2$,$\cot x = 0$,so $[0] + [0] = 0$. For all other $x \in (0, \pi)$,$\cot x$ is not an integer. Thus,$[\cot x] + [-\cot x] = -1$ almost everywhere in $(0, \pi)$. $2I = \int_{0}^{\pi} (-1) dx = -\pi$. Therefore,$I = -\frac{\pi}{2}$.

$\int_{0}^{\pi} [\cot x] dx = $

Similar Questions

The value of the integral $\int_{-2}^{2} \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}} \, dx$ (where $[x]$ denotes the greatest integer less than or equal to $x$) is

The option$(s)$ with the values of $a$ and $L$ that satisfy the following equation is(are) $\frac{\int_0^{4 \pi} e^t(\sin^6 at + \cos^4 at) dt}{\int_0^{\pi} e^t(\sin^6 at + \cos^4 at) dt} = L$.

$\int_0^{\pi /2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx = $

Let $I_n = \int_0^{\pi / 2} x^n \cos x \, dx$,where $n$ is a non-negative integer. Then,$\sum_{n=2}^{\infty} \left( \frac{I_n}{n!} + \frac{I_{n-2}}{(n-2)!} \right)$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module