int_{0}^{frac{pi}{2}} frac{sqrt[3]{sec x}}{sq | vedclass

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

Question

(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{(\sec x)^{1/3}}{(\sec x)^{1/3} + (\operatorname{cosec} x)^{1/3}} dx \quad ...(1)$ Using the property $\int

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{(\sec x)^{1/3}}{(\sec x)^{1/3} + (\operatorname{cosec} x)^{1/3}} dx \quad ...(1)$ Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have: $I = \int_{0}^{\frac{\pi}{2}} \frac{(\sec(\frac{\pi}{2}-x))^{1/3}}{(\sec(\frac{\pi}{2}-x))^{1/3} + (\operatorname{cosec}(\frac{\pi}{2}-x))^{1/3}} dx$ Since $\sec(\frac{\pi}{2}-x) = \operatorname{cosec} x$ and $\operatorname{cosec}(\frac{\pi}{2}-x) = \sec x$,we get: $I = \int_{0}^{\frac{\pi}{2}} \frac{(\operatorname{cosec} x)^{1/3}}{(\operatorname{cosec} x)^{1/3} + (\sec x)^{1/3}} dx \quad ...(2)$ Adding equation $(1)$ and $(2)$: $2I = \int_{0}^{\frac{\pi}{2}} \frac{(\sec x)^{1/3} + (\operatorname{cosec} x)^{1/3}}{(\sec x)^{1/3} + (\operatorname{cosec} x)^{1/3}} dx$ $2I = \int_{0}^{\frac{\pi}{2}} 1 dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}$ Therefore,$I = \frac{\pi}{4}$.

$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt[3]{\sec x}}{\sqrt[3]{\sec x}+\sqrt[3]{\operatorname{cosec} x}} d x=$

Similar Questions

$\int_0^1 \frac{\log x}{\sqrt{1 - x^2}} \, dx = $

The value of the integral $I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\tan 2x}}$ is:

$\frac{3}{25} \int_0^{25 \pi} \sqrt{|\cos x - \cos^3 x|} \, dx =$

If $f(a+b-x)=f(x)$,then $\int_a^b x f(x) d x=$ . . . . . . .

Evaluate $\int_{-1}^{1} \sin^{5} x \cos^{4} x \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module