int_{-pi / 2}^{pi / 2} frac{cos ^{2} x}{1+3^{ | vedclass

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

Question

(A) ધારો કે $I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+3^{x}} dx$  --- $(1)$ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ નો ઉપયો

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) ધારો કે $I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+3^{x}} dx$  --- $(1)$ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ નો ઉપયોગ કરતા:$I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2}(-x)}{1+3^{-x}} dx = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+\frac{1}{3^{x}}} dx = \int_{-\pi / 2}^{\pi / 2} \frac{3^{x} \cos ^{2} x}{3^{x}+1} dx$ --- $(2)$$(1)$ અને $(2)$ નો સરવાળો કરતા:$2I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x + 3^{x} \cos ^{2} x}{1+3^{x}} dx$$2I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x(1+3^{x})}{1+3^{x}} dx = \int_{-\pi / 2}^{\pi / 2} \cos ^{2} x dx$$\cos^{2} x$ એ યુગ્મ વિધેય હોવાથી:$2I = 2 \int_{0}^{\pi / 2} \cos ^{2} x dx$$I = \int_{0}^{\pi / 2} \frac{1+\cos 2x}{2} dx = \frac{1}{2} [x + \frac{\sin 2x}{2}]_{0}^{\pi / 2}$$I = \frac{1}{2} [(\frac{\pi}{2} + 0) - (0 + 0)] = \frac{\pi}{4}$

$\int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+3^{x}} d x$ નું મૂલ્ય શોધો.

Similar Questions

જો $g(x) = \int_0^x \cos^4 t \,dt$ હોય, તો $g(x+\pi)$ ની કિંમત શું થાય?

$\int_{2}^{4} \frac{\log(x^2)}{\log(x^2) + \log(36 - 12x + x^2)} \, dx = $

$\int_{0}^{\pi} x \sin^{3} x \, dx$ નું મૂલ્ય શું છે?

$\int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{5 / 2} d x=$

સંકલન $\int_{-\pi / 2}^{\pi / 2} \frac{\sin^2 x}{1+e^x} \, dx$ નું મૂલ્ય શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module