int cos x sqrt{4 - sin^2 x} ; dx = | vedclass

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Question

(B) Let $I = \int \cos x \sqrt{4 - \sin^2 x} \; dx$. Substitute $\sin x = t$,then $\cos x \; dx = dt$. The integral becomes $I = \int \sqrt{4 - t^2} \

Vedclass · Accepted Answer

$\frac{1}{2} \sin x \sqrt{4 - \sin^2 x} + 2 \sin^{-1} \left( \frac{1}{2} \sin x \right) + c$

Vedclass · Answer

(B) Let $I = \int \cos x \sqrt{4 - \sin^2 x} \; dx$. Substitute $\sin x = t$,then $\cos x \; dx = dt$. The integral becomes $I = \int \sqrt{4 - t^2} \; dt = \int \sqrt{2^2 - t^2} \; dt$. Using the standard formula $\int \sqrt{a^2 - x^2} \; dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) + c$,we get: $I = \frac{t}{2} \sqrt{4 - t^2} + \frac{4}{2} \sin^{-1} \left( \frac{t}{2} \right) + c$. Substituting $t = \sin x$ back into the expression: $I = \frac{1}{2} \sin x \sqrt{4 - \sin^2 x} + 2 \sin^{-1} \left( \frac{\sin x}{2} \right) + c$.

$\int \cos x \sqrt{4 - \sin^2 x} \; dx = $

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