int_{-frac{pi}{2}}^{frac{pi}{2}} frac{sin ^2 | vedclass

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Question

(A) ધારો કે $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^x} \,d x$ ... $(i)$ ગુણધર્મ $\int_{a}^{b} f(x) \,d x = \int_{a}^{b} f(a+b-

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

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

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^x} \,d x$ નું મૂલ્ય શોધો.

Similar Questions

ધારો કે $u = \int\limits_0^1 {\frac{{\ln (x + 1)}}{{{x^2} + 1}}} \,dx$ અને $v = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin 2x)} \,dx$,તો:

નિશ્ચિત સંકલન $\int\limits_0^{\frac{\pi }{2}} {\sqrt {\tan x} \,dx} $ નું મૂલ્ય છે

$\int_0^a f(x) \, dx = $

$\int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + {e^x}}}\,dx = } $

$\int_{0}^{1} \sin \left( 2 \tan^{-1} \sqrt{\frac{1+x}{1-x}} \right) \, dx = $

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