int_0^{pi /2} frac{cos x}{1 + cos x + sin x} | vedclass

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

Question

(C) ધારો કે $I = \int_0^{\pi /2} \frac{\cos x}{1 + \cos x + \sin x} \,dx$. ગુણધર્મ $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$ નો ઉપયોગ કરતા: $I = \in

Vedclass · Accepted Answer

$\frac{\pi }{4} - \frac{1}{2}\log 2$

Vedclass · Answer

(C) ધારો કે $I = \int_0^{\pi /2} \frac{\cos x}{1 + \cos x + \sin x} \,dx$. ગુણધર્મ $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$ નો ઉપયોગ કરતા: $I = \int_0^{\pi /2} \frac{\cos(\pi/2 - x)}{1 + \cos(\pi/2 - x) + \sin(\pi/2 - x)} \,dx = \int_0^{\pi /2} \frac{\sin x}{1 + \sin x + \cos x} \,dx$. બંને સમીકરણોનો સરવાળો કરતા: $2I = \int_0^{\pi /2} \frac{\cos x + \sin x}{1 + \cos x + \sin x} \,dx$. આને આ રીતે લખી શકાય: $2I = \int_0^{\pi /2} \frac{(1 + \cos x + \sin x) - 1}{1 + \cos x + \sin x} \,dx = \int_0^{\pi /2} 1 \,dx - \int_0^{\pi /2} \frac{1}{1 + \cos x + \sin x} \,dx$. $2I = \frac{\pi}{2} - \int_0^{\pi /2} \frac{1}{2\cos^2(x/2) + 2\sin(x/2)\cos(x/2)} \,dx = \frac{\pi}{2} - \frac{1}{2} \int_0^{\pi /2} \sec^2(x/2) \frac{1}{1 + 	an(x/2)} \,dx$. ધારો કે $u = 	an(x/2)$,તો $du = \frac{1}{2} \sec^2(x/2) \,dx$. સીમાઓ: $x=0 \implies u=0$,$x=\pi/2 \implies u=1$. $2I = \frac{\pi}{2} - \int_0^1 \frac{1}{1+u} \,du = \frac{\pi}{2} - [\ln(1+u)]_0^1 = \frac{\pi}{2} - \ln 2$. $I = \frac{\pi}{4} - \frac{1}{2} \ln 2$.

$\int_0^{\pi /2} \frac{\cos x}{1 + \cos x + \sin x} \,dx = $

Similar Questions

$\int_{0}^{1} \frac{8 \log(1+x)}{1+x^{2}} dx = $

$\int_0^{\pi / 2} \frac{(\cos x)^{\sin x}}{(\cos x)^{\sin x}+(\sin x)^{\cos x}} d x$ નું મૂલ્ય શું છે?

$\int_0^{\pi / 4} \log (1+\tan x) d x=$

$\int_{-1}^{1} \log \left( \frac{1+x}{1-x} \right) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module