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

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(D) 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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) 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 $\sin x = t$,then $\cos x \, dx = dt$. $I = \int \sqrt{4 - 9t^2} \, dt = \int \sqrt{2^2 - (3t)^2} \, dt$. Using the formula $\int \sqrt{a^2 - u^2} \, du = \frac{u}{2} \sqrt{a^2 - u^2} + \frac{a^2}{2} \sin ^{-1}\left(\frac{u}{a}\right) + C$,where $u = 3t$ and $du = 3 \, dt$ (so $dt = \frac{du}{3}$): $I = \frac{1}{3} \int \sqrt{2^2 - u^2} \, du = \frac{1}{3} \left[ \frac{u}{2} \sqrt{4 - u^2} + \frac{4}{2} \sin ^{-1}\left(\frac{u}{2}\right) \right] + C$. Substituting $u = 3 \sin x$: $I = \frac{1}{3} \left[ \frac{3 \sin x}{2} \sqrt{4 - 9 \sin ^2 x} + 2 \sin ^{-1}\left(\frac{3 \sin x}{2}\right) \right] + C$. $I = \frac{1}{2} \sin x \sqrt{4 - 9 \sin ^2 x} + \frac{2}{3} \sin ^{-1}\left(\frac{3 \sin x}{2}\right) + C$.

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

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