int_{frac{pi}{18}}^{frac{4pi}{9}} frac{2sqrt{ | vedclass

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

Question

(C) ધારો કે $I = \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{2\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ . . . $(i)$ ગુણધર્મ $\int_{a}^{b} f(x

Vedclass · Accepted Answer

$\frac{7\pi}{18}$

Vedclass · Answer

(C) ધારો કે $I = \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{2\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ . . . $(i)$ ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ નો ઉપયોગ કરતા,જ્યાં $a = \frac{\pi}{18}$ અને $b = \frac{4\pi}{9}$,આપણને $a+b = \frac{\pi}{18} + \frac{8\pi}{18} = \frac{9\pi}{18} = \frac{\pi}{2}$ મળે છે. તેથી,$I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} dx$ $I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ . . . $(ii)$ $(i)$ અને $(ii)$ નો સરવાળો કરતા: $2I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \left( \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) dx$ $2I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} 1 dx = 2 [x]_{\frac{\pi}{18}}^{\frac{4\pi}{9}}$ $2I = 2 \left( \frac{4\pi}{9} - \frac{\pi}{18} \right) = 2 \left( \frac{8\pi - \pi}{18} \right) = 2 \left( \frac{7\pi}{18} \right) = \frac{7\pi}{9}$ $I = \frac{7\pi}{18}$

$\int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{2\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = . . . . . .$

Similar Questions

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

$\int_{0}^{\pi /2} \frac{\sin^{2/3} x}{\sin^{2/3} x + \cos^{2/3} x} dx$ નું મૂલ્ય શોધો.

સંકલન $\int_{0}^{1} x \cot^{-1}(1 - x^2 + x^4) dx$ નું મૂલ્ય શું છે?

$\int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module