int_{-frac{pi}{4}}^{frac{pi}{4}} sin^{2} x , | vedclass

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

Question

(A) અહીં $f(x) = \sin^{2} x$ એ યુગ્મ વિધેય છે કારણ કે $f(-x) = \sin^{2}(-x) = (-\sin x)^{2} = \sin^{2} x = f(x)$.તેથી,નિશ્ચિત સંકલનના ગુણધર્મ $\int_{-

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) અહીં $f(x) = \sin^{2} x$ એ યુગ્મ વિધેય છે કારણ કે $f(-x) = \sin^{2}(-x) = (-\sin x)^{2} = \sin^{2} x = f(x)$.તેથી,નિશ્ચિત સંકલનના ગુણધર્મ $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ મુજબ,આપણને મળે છે:$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^{2} x \, dx = 2 \int_{0}^{\frac{\pi}{4}} \sin^{2} x \, dx$નિત્યસમ $\sin^{2} x = \frac{1 - \cos 2x}{2}$ નો ઉપયોગ કરતા:$= 2 \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos 2x}{2} \, dx = \int_{0}^{\frac{\pi}{4}} (1 - \cos 2x) \, dx$$= \left[ x - \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{4}}$$= \left( \frac{\pi}{4} - \frac{\sin(\frac{\pi}{2})}{2} \right) - (0 - 0)$$= \frac{\pi}{4} - \frac{1}{2} = \frac{\pi - 2}{4}$

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^{2} x \, dx$ ની કિંમત શોધો.

Similar Questions

$\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{\tan x}{\tan x + \cot x} \, dx =$

$\int_{-\pi / 15}^{\pi / 15} \frac{\cos 5 x}{1+e^{5 x}} d x=$

$\int_0^{\frac{\pi}{2}} \left( \frac{\sqrt[n]{\sec x}}{\sqrt[n]{\sec x} + \sqrt[n]{\operatorname{cosec} x}} \right) dx = $

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module