int_{,pi /6}^{,pi /3} {,frac{{dx}}{{1 + sqrt | vedclass

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

Question

(C) Let $I = \int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{\cot x}}$.We can write $\sqrt{\cot x} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}}$,so $I = \int_{\pi

Vedclass · Accepted Answer

$\pi /12$

Vedclass · Answer

(C) Let $I = \int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{\cot x}}$.We can write $\sqrt{\cot x} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}}$,so $I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ ....$(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a = \pi/6$ and $b = \pi/3$,we have $a+b = \pi/2$.Thus,$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\sin(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} dx = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ ....(ii)Adding $(i)$ and (ii):$2I = \int_{\pi /6}^{\pi /3} \left( \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} ight) dx = \int_{\pi /6}^{\pi /3} 1 dx$$2I = [x]_{\pi /6}^{\pi /3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$Therefore,$I = \frac{1}{2} 	imes \frac{\pi}{6} = \frac{\pi}{12}$.

$\int_{\,\pi /6}^{\,\pi /3} {\,\frac{{dx}}{{1 + \sqrt {\cot x} }}} $ is equal to:

Similar Questions

The value of $I = \int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)} + e^{\tan^{-1}(\cos x)}} dx$ is

If $f(5-x)=f(x)$ and $\int_2^3 f(x) dx=2$,then $\int_2^3 x f(x) dx=$

If $f(x)$ is an odd function of $x,$ then $\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {f(\cos x)\,dx} $ is equal to

Let $f(x) = \int\limits_1^x \frac{\tan^{-1} t}{t} dt$ for $x > 0$. Then $f(e^2) - f\left(\frac{1}{e^2}\right)$ is

If $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$,then $[I] = \ldots$. Here,$[.]$ denotes the greatest integer function.

Vedclass Test Series

Exam Paper Generator

Online Exam Module