If int_0^{frac{pi}{2}} frac{cot x}{cot x+oper | vedclass

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

Question

(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{\cot x}{\cot x+\operatorname{cosec} x} d x$. Simplifying the integrand: $\frac{\cot x}{\cot x+\operatorname{

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{\cot x}{\cot x+\operatorname{cosec} x} d x$. Simplifying the integrand: $\frac{\cot x}{\cot x+\operatorname{cosec} x} = \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x} + \frac{1}{\sin x}} = \frac{\cos x}{\cos x+1}$. Now,$I = \int_0^{\frac{\pi}{2}} \frac{\cos x}{1+\cos x} d x = \int_0^{\frac{\pi}{2}} \left( \frac{1+\cos x - 1}{1+\cos x} ight) d x = \int_0^{\frac{\pi}{2}} \left( 1 - \frac{1}{1+\cos x} ight) d x$. Using the identity $1+\cos x = 2\cos^2 \frac{x}{2}$,we get $I = \int_0^{\frac{\pi}{2}} \left( 1 - \frac{1}{2}\sec^2 \frac{x}{2} ight) d x$. Integrating,we get $I = \left[ x - 	an \frac{x}{2} ight]_0^{\frac{\pi}{2}} = \left( \frac{\pi}{2} - 	an \frac{\pi}{4} ight) - (0 - 	an 0) = \frac{\pi}{2} - 1$. We can write this as $\frac{1}{2}(\pi - 2)$. Comparing with $m(\pi+n)$,we get $m = \frac{1}{2}$ and $n = -2$. Therefore,$m \cdot n = \frac{1}{2} \cdot (-2) = -1$.

If $\int_0^{\frac{\pi}{2}} \frac{\cot x}{\cot x+\operatorname{cosec} x} d x=m(\pi+n)$,then $(m \cdot n)$ equals

Similar Questions

$\int_{0}^{\pi} \sin x \, dx = $ . . . . . . .

Evaluate the definite integral $\int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$.

$\int_{1}^{x} \frac{\log(x^2)}{x} \, dx = $

$\int_{0}^{\pi /2} \{ x - [\sin x] \} \,dx$ is equal to

$\int_0^a x^2 (a^2 - x^2)^{3/2} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module