int_{0}^{frac{pi}{2}} frac{sqrt[7]{sin x}}{sq | vedclass

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

Question

(C) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}} dx$ ... $(1)$Using the property $\int_{0}^{a} f(x) dx

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}} dx$ ... $(1)$Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:$I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin(\frac{\pi}{2}-x)}}{\sqrt[7]{\sin(\frac{\pi}{2}-x)}+\sqrt[7]{\cos(\frac{\pi}{2}-x)}} dx$$I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\cos x}}{\sqrt[7]{\cos x}+\sqrt[7]{\sin x}} dx$ ... $(2)$Adding equations $(1)$ and $(2)$:$2I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x} + \sqrt[7]{\cos x}}{\sqrt[7]{\sin x} + \sqrt[7]{\cos x}} dx$$2I = \int_{0}^{\frac{\pi}{2}} 1 dx$$2I = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}$$I = \frac{\pi}{4}$

$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}} dx =$

Similar Questions

$\int_{-3}^{3} \frac{x^2 \sin x}{1 + x^6} \, dx = $

If $n$ is a positive integer and $[x]$ is the greatest integer not exceeding $x$,then $\int_0^n {\{x - [x]\} \,dx}$ equals

$\int_{-\pi / 8}^{\pi / 8} \frac{\sin ^4(4 x)}{1+e^{4 x}} d x=$

The value of the definite integral $\int_{19}^{37} (\{x\}^2 + 3 \sin(2\pi x)) \, dx$,where $\{x\}$ denotes the fractional part function.

Evaluate the definite integral: $\int_{0}^{\infty} \frac{x}{(1 + x)(1 + x^2)} \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module