int sqrt{1 + sin x} , dx ની કિંમત શું થાય? | vedclass

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Question

(A) આપણે જાણીએ છીએ કે $1 + \sin x = \cos^2(x/2) + \sin^2(x/2) + 2\sin(x/2)\cos(x/2) = (\cos(x/2) + \sin(x/2))^2$ થાય.તેથી, $\sqrt{1 + \sin x} = |\cos(

Vedclass · Accepted Answer

$2\sqrt{2} \sin \left( \frac{x}{2} - \frac{\pi}{4} \right) + c$

Vedclass · Answer

(A) આપણે જાણીએ છીએ કે $1 + \sin x = \cos^2(x/2) + \sin^2(x/2) + 2\sin(x/2)\cos(x/2) = (\cos(x/2) + \sin(x/2))^2$ થાય.તેથી, $\sqrt{1 + \sin x} = |\cos(x/2) + \sin(x/2)|$ થાય.જે અંતરાલમાં $\cos(x/2) + \sin(x/2) > 0$ હોય તે લેતા, $\int (\cos(x/2) + \sin(x/2)) \, dx$ મળે.$= 2\sin(x/2) - 2\cos(x/2) + c$ મળે.નિત્યસમ $\sin(A-B) = \sin A \cos B - \cos A \sin B$ નો ઉપયોગ કરતા, આપણે $2\sin(x/2) - 2\cos(x/2) = 2\sqrt{2} \left( \frac{1}{\sqrt{2}}\sin(x/2) - \frac{1}{\sqrt{2}}\cos(x/2) \right) = 2\sqrt{2} \sin(x/2 - \pi/4) + c$ લખી શકીએ.

$\int \sqrt{1 + \sin x} \, dx$ ની કિંમત શું થાય?

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