int_{-1 / 2}^{1 / 2} cos ^{-1} x , dx નું મૂલ | vedclass

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Question

(B) આપણી પાસે છે,$I = \int_{-1 / 2}^{1 / 2} \cos ^{-1} x \, dx$. ગુણધર્મ $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$ નો ઉપયોગ કરતા: $I = \int_{-1 / 2}

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(B) આપણી પાસે છે,$I = \int_{-1 / 2}^{1 / 2} \cos ^{-1} x \, dx$. ગુણધર્મ $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$ નો ઉપયોગ કરતા: $I = \int_{-1 / 2}^{1 / 2} (\frac{\pi}{2} - \sin^{-1} x) \, dx$. $I = \int_{-1 / 2}^{1 / 2} \frac{\pi}{2} \, dx - \int_{-1 / 2}^{1 / 2} \sin^{-1} x \, dx$. $\sin^{-1} x$ એ અયુગ્મ વિધેય હોવાથી,સંકલન $\int_{-a}^{a} \sin^{-1} x \, dx = 0$ થાય. તેથી,$I = \frac{\pi}{2} [x]_{-1 / 2}^{1 / 2} - 0$. $I = \frac{\pi}{2} (\frac{1}{2} - (- \frac{1}{2})) = \frac{\pi}{2} (1) = \frac{\pi}{2}$.

$\int_{-1 / 2}^{1 / 2} \cos ^{-1} x \, dx$ નું મૂલ્ય શોધો.

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