int_{pi /6}^{pi /3} frac{dx}{1 + sqrt{tan x}} | vedclass

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

Question

(A) माना $I = \int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{	an x}}$.$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$

Vedclass · Accepted Answer

$\pi /12$

Vedclass · Answer

(A) माना $I = \int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{	an x}}$.$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ ...$(i)$गुणधर्म $\int_a^b f(x) dx = \int_a^b f(a + b - x) dx$ का उपयोग करने पर,हमें प्राप्त होता है:$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos(\pi/2 - x)}}{\sqrt{\cos(\pi/2 - x)} + \sqrt{\sin(\pi/2 - x)}} dx$$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ ...(ii)$(i)$ और $(ii)$ को जोड़ने पर:$2I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$$2I = \int_{\pi /6}^{\pi /3} 1 dx$$2I = [x]_{\pi /6}^{\pi /3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$$I = \frac{\pi}{12}$.

$\int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{\tan x}} = $

Similar Questions

$\int_0^{\frac{\pi}{4}} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x=$

यदि $\int_{0}^{1} 4 \cot^{-1}(1-x+x^{2}) dx = a \tan^{-1}(2) - b \log_{e}(5)$, जहाँ $a, b \in N$, तो $(2a+b)$ का मान ज्ञात कीजिए:

समाकल $\int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$ का मान किसके बराबर है?

यदि $F(x) = f(x) + f\left(\frac{1}{x}\right)$,जहाँ $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt$ है,तो $F(e) = $

$\int_0^1 \tan^{-1}(1-x+x^2) dx$ का मान है

Vedclass Test Series

Exam Paper Generator

Online Exam Module