int sqrt{4 cos ^2 x - 5 sin ^2 x} cos x , dx | vedclass

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Question

(A) Let $I = \int \sqrt{4 \cos ^2 x - 5 \sin ^2 x} \cos x \, dx$. Using the identity $\cos ^2 x = 1 - \sin ^2 x$, we get: $I = \int \sqrt{4(1 - \sin ^

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$\frac{1}{2} \sin x \sqrt{4 - 9 \sin ^2 x} + \frac{2}{3} \sin ^{-1}\left(\frac{3 \sin x}{2}\right) + c$

Vedclass · Answer

(A) Let $I = \int \sqrt{4 \cos ^2 x - 5 \sin ^2 x} \cos x \, dx$. Using the identity $\cos ^2 x = 1 - \sin ^2 x$, we get: $I = \int \sqrt{4(1 - \sin ^2 x) - 5 \sin ^2 x} \cos x \, dx = \int \sqrt{4 - 9 \sin ^2 x} \cos x \, dx$. Let $3 \sin x = t$. Then $3 \cos x \, dx = dt$, which implies $\cos x \, dx = \frac{1}{3} dt$. Substituting these into the integral: $I = \int \sqrt{4 - t^2} \cdot \frac{1}{3} dt = \frac{1}{3} \int \sqrt{2^2 - t^2} dt$. Using the standard formula $\int \sqrt{a^2 - x^2} dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right) + c$: $I = \frac{1}{3} \left[ \frac{t}{2} \sqrt{4 - t^2} + \frac{4}{2} \sin ^{-1}\left(\frac{t}{2}\right) \right] + c$. Substituting $t = 3 \sin x$ back: $I = \frac{1}{3} \left[ \frac{3 \sin x}{2} \sqrt{4 - 9 \sin ^2 x} + 2 \sin ^{-1}\left(\frac{3 \sin x}{2}\right) \right] + c = \frac{1}{2} \sin x \sqrt{4 - 9 \sin ^2 x} + \frac{2}{3} \sin ^{-1}\left(\frac{3 \sin x}{2}\right) + c$.

$\int \sqrt{4 \cos ^2 x - 5 \sin ^2 x} \cos x \, dx =$

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