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

Prove that $\int_{-1}^{1} x^{17} \cos^{4} x \, dx = 0$.

The value of $\int_{-\pi/4}^{\pi/4} \sin^{103} x \cdot \cos^{101} x \, dx$ is

The function $F(x) = \int_0^x \log \left( \frac{1 - t}{1 + t} \right) \,dt$ is

If $f : R \rightarrow R$ is a continuous function satisfying $\int \limits_0^{\pi / 2} f(\sin 2x) \cdot \sin x \, dx + \alpha \int \limits_0^{\pi / 4} f(\cos 2x) \cdot \cos x \, dx = 0$,then $\alpha$ is equal to

$\int_5^9 \frac{\log_3 x^2}{\log_3 x^2 + \log_3(588 - 84x + 3x^2)} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module