int_0^{pi /2} frac{sqrt{cot x}}{sqrt{cot x} + | vedclass

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

Question

(C) माना $I = \int_0^{\pi /2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{	an x}} \, dx$ ..... $(i)$गुणधर्म $\int_0^a f(x) \, dx = \int_0^a f(a-x) \,

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) माना $I = \int_0^{\pi /2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{	an x}} \, dx$ ..... $(i)$गुणधर्म $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ का उपयोग करने पर:$I = \int_0^{\pi /2} \frac{\sqrt{\cot(\pi/2 - x)}}{\sqrt{\cot(\pi/2 - x)} + \sqrt{	an(\pi/2 - x)}} \, dx$चूंकि $\cot(\pi/2 - x) = 	an x$ और $	an(\pi/2 - x) = \cot x$,इसलिए:$I = \int_0^{\pi /2} \frac{\sqrt{	an x}}{\sqrt{	an x} + \sqrt{\cot x}} \, dx$ ..... $(ii)$$(i)$ और $(ii)$ को जोड़ने पर:$2I = \int_0^{\pi /2} \frac{\sqrt{\cot x} + \sqrt{	an x}}{\sqrt{\cot x} + \sqrt{	an x}} \, dx$$2I = \int_0^{\pi /2} 1 \, dx$$2I = [x]_0^{\pi /2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$I = \frac{\pi}{4}$

$\int_0^{\pi /2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} \, dx = $

Similar Questions

$\int\limits_{ - 1}^1 {\frac{{{x^3} + |x| + 1}}{{{x^2} + 2|x| + 1}}} dx = a \ln 2 + b$,तो:

$\int_{-100}^{100} \frac{x+x^3+x^5}{1+x^2+x^4+x^6} dx$ का मान ज्ञात कीजिए।

$\int_{-5}^{5} \left[ \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} \right] dx = $

निश्चित समाकलनों के गुणों का उपयोग करके,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{7} x \, dx$ का मान ज्ञात कीजिए।

निश्चित समाकलन $\int\limits_0^{\frac{\pi }{2}} {\sqrt {\tan x} \,dx} $ का मान है

Vedclass Test Series

Exam Paper Generator

Online Exam Module