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) माना $I = \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{2\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ . . . $(i)$ गुणधर्म $\int_{a}^{b} f(x) d

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) माना $I = \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{2\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ . . . $(i)$ गुणधर्म $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ का उपयोग करने पर,जहाँ $a = \frac{\pi}{18}$ और $b = \frac{4\pi}{9}$,हमें $a+b = \frac{\pi}{18} + \frac{8\pi}{18} = \frac{9\pi}{18} = \frac{\pi}{2}$ प्राप्त होता है। अतः,$I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} dx$ $I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ . . . $(ii)$ $(i)$ और $(ii)$ को जोड़ने पर: $2I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \left( \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) dx$ $2I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} 1 dx = 2 [x]_{\frac{\pi}{18}}^{\frac{4\pi}{9}}$ $2I = 2 \left( \frac{4\pi}{9} - \frac{\pi}{18} \right) = 2 \left( \frac{8\pi - \pi}{18} \right) = 2 \left( \frac{7\pi}{18} \right) = \frac{7\pi}{9}$ $I = \frac{7\pi}{18}$

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

Similar Questions

$\frac{8}{\pi} \int \limits_0^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}} dx$ का मान $.............$ है।

$\int_{0}^{\pi} \log (1+\cos x) d x$ का मान ज्ञात कीजिए।

$\int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} (|\cos t| \sin t + |\sin t| \cos t) dt =$

$g(\alpha)$ के लिए निम्नलिखित में से कौन सा कथन गलत है,जहाँ $\alpha \in R$ और $g(\alpha)=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^{\alpha} x}{\cos^{\alpha} x+\sin^{\alpha} x} dx$ है?

$\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module