જો I = frac{2}{pi} int_{-pi / 4}^{pi / 4} fra | vedclass

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

Question

(B) ધારો કે $I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$.ગુણધર્મ $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ ન

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) ધારો કે $I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$.ગુણધર્મ $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ નો ઉપયોગ કરતા:$I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{-\sin x})(2 - \cos 2x)}$.$I$ માટેના બંને પદોનો સરવાળો કરતા:$2I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{1}{2 - \cos 2x} \left( \frac{1}{1 + e^{\sin x}} + \frac{e^{\sin x}}{1 + e^{\sin x}} ight) dx = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{2 - \cos 2x}$.વિધેય યુગ્મ હોવાથી,$I = \frac{2}{\pi} \int_0^{\pi / 4} \frac{dx}{2 - \cos 2x} = \frac{2}{\pi} \int_0^{\pi / 4} \frac{\sec^2 x dx}{2(1 + 	an^2 x) - (1 - 	an^2 x)} = \frac{2}{\pi} \int_0^{\pi / 4} \frac{\sec^2 x dx}{1 + 3 	an^2 x}$.ધારો કે $u = \sqrt{3} 	an x$,તો $du = \sqrt{3} \sec^2 x dx$.$I = \frac{2}{\pi \sqrt{3}} \int_0^{\sqrt{3}} \frac{du}{1 + u^2} = \frac{2}{\pi \sqrt{3}} [	an^{-1} u]_0^{\sqrt{3}} = \frac{2}{\pi \sqrt{3}} \cdot \frac{\pi}{3} = \frac{2}{3\sqrt{3}}$.તેથી,$27 I^2 = 27 \cdot \frac{4}{27} = 4$.

જો $I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$ હોય,તો $27 I^2$ ની કિંમત . . . . . . . . થાય.

Similar Questions

$\int_0^{\frac{\pi}{2}} \frac{dx}{1+(\cot x)^{101}} = $

જો $\int_0^{\frac{\pi}{2}} \log \cos x \, dx = \frac{\pi}{2} \log \left(\frac{1}{2}\right)$ હોય,તો $\int_0^{\frac{\pi}{2}} \log \sec x \, dx = $

$\int_0^{\pi /2} |\sin x - \cos x| \, dx = $

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

સંકલન $\int_{-\pi}^{\pi} \frac{\cos^2 x}{1+a^x} dx$ નું મૂલ્ય,જ્યાં $a > 0$,શું છે?

Vedclass Test Series

Exam Paper Generator

Online Exam Module