int_0^{frac{pi}{2}} left( frac{sqrt[n]{sec x} | vedclass

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

Question

(C) Let $I = \int_0^{\frac{\pi}{2}} \frac{(\sec x)^{1/n}}{(\sec x)^{1/n} + (\operatorname{cosec} x)^{1/n}} dx \dots (i)$ Using the property $\int_0^a

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) Let $I = \int_0^{\frac{\pi}{2}} \frac{(\sec x)^{1/n}}{(\sec x)^{1/n} + (\operatorname{cosec} x)^{1/n}} dx \dots (i)$ Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$: $I = \int_0^{\frac{\pi}{2}} \frac{(\sec(\frac{\pi}{2}-x))^{1/n}}{(\sec(\frac{\pi}{2}-x))^{1/n} + (\operatorname{cosec}(\frac{\pi}{2}-x))^{1/n}} dx$ $I = \int_0^{\frac{\pi}{2}} \frac{(\operatorname{cosec} x)^{1/n}}{(\operatorname{cosec} x)^{1/n} + (\sec x)^{1/n}} dx \dots (ii)$ Adding $(i)$ and $(ii)$: $2I = \int_0^{\frac{\pi}{2}} \frac{(\sec x)^{1/n} + (\operatorname{cosec} x)^{1/n}}{(\sec x)^{1/n} + (\operatorname{cosec} x)^{1/n}} dx$ $2I = \int_0^{\frac{\pi}{2}} 1 dx$ $2I = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$ $I = \frac{\pi}{4}$

$\int_0^{\frac{\pi}{2}} \left( \frac{\sqrt[n]{\sec x}}{\sqrt[n]{\sec x} + \sqrt[n]{\operatorname{cosec} x}} \right) dx = $

Similar Questions

If the value of $\int_{0}^{\pi/2} \sin^{4}(x) \cdot \cos^{2}(x) dx = \frac{\pi}{32}$,then the value of $\int_{0}^{\pi/2} \cos^{4}(x) \cdot \sin^{2}(x) dx$ is:

$\int_{0}^{\pi} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} = $

$\int_0^{\pi /2} {x\cot x\,dx} $ equals

If $\int_0^{10} f(x) d x=5$,then $\sum_{k=1}^{10} \int_0^1 f(k-1+x) d x=$

If $[ \cdot ]$ represents the greatest integer function,then $\int_{-1}^1 (x[1+\sin(\pi x)]+1) dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module