int_{-pi/2}^{pi/2} frac{sin^2 x}{1 + 2^x} dx | vedclass

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

Question

(C) ધારો કે $I = \int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1 + 2^x} dx$  $(i)$ ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ નો ઉપયોગ કરતા: $I

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) ધારો કે $I = \int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1 + 2^x} dx$  $(i)$ ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ નો ઉપયોગ કરતા: $I = \int_{-\pi/2}^{\pi/2} \frac{\sin^2(-x)}{1 + 2^{-x}} dx = \int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1 + \frac{1}{2^x}} dx = \int_{-\pi/2}^{\pi/2} \frac{2^x \sin^2 x}{2^x + 1} dx$  $(ii)$ $(i)$ અને $(ii)$ નો સરવાળો કરતા: $2I = \int_{-\pi/2}^{\pi/2} \frac{\sin^2 x + 2^x \sin^2 x}{1 + 2^x} dx = \int_{-\pi/2}^{\pi/2} \sin^2 x dx$ $\sin^2 x$ એ યુગ્મ વિધેય હોવાથી,$2I = 2 \int_{0}^{\pi/2} \sin^2 x dx$ $I = \int_{0}^{\pi/2} \sin^2 x dx = \int_{0}^{\pi/2} \frac{1 - \cos 2x}{2} dx = \left[ \frac{x}{2} - \frac{\sin 2x}{4} \right]_{0}^{\pi/2} = \frac{\pi}{4} - 0 = \frac{\pi}{4}$

$\int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1 + 2^x} dx = \dots$

Similar Questions

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\tan^4 x} dx = $ . . . . . . .

$\int_0^{\infty} (x^{12} + x^{-12}) \frac{\log x}{x} dx =$

$\int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^{x}-1}{e^{x}+1}\right) dx=$

જો $f(\alpha) = \int_{1}^{\alpha} \frac{\log_{10} t}{1+t} dt, \alpha > 0$ હોય,તો $f(e^{3}) + f(e^{-3})$ ની કિંમત શોધો.

જો $\int_0^{\pi / 2} \tan ^{14}\left(\frac{x}{2}\right) d x=2\left[\sum_{n=1}^7 f(n)-\frac{\pi}{4}\right]$ હોય, તો $f(n)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module