If int frac{2 sin 2x - 3 cos x}{2 sin^2 x - 3 | vedclass

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(C) Given the integral $I = \int \frac{2 \sin 2x - 3 \cos x}{2 \sin^2 x - 3 \sin x + 4} dx$. Using the identity $\sin 2x = 2 \sin x \cos x$, we rewrit

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$\log \left(\frac{3}{4}\right)$

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(C) Given the integral $I = \int \frac{2 \sin 2x - 3 \cos x}{2 \sin^2 x - 3 \sin x + 4} dx$. Using the identity $\sin 2x = 2 \sin x \cos x$, we rewrite the numerator: $I = \int \frac{4 \sin x \cos x - 3 \cos x}{2 \sin^2 x - 3 \sin x + 4} dx = \int \frac{(4 \sin x - 3) \cos x}{2 \sin^2 x - 3 \sin x + 4} dx$. Let $\sin x = t$, then $\cos x dx = dt$. The integral becomes $I = \int \frac{4t - 3}{2t^2 - 3t + 4} dt$. Let $u = 2t^2 - 3t + 4$, then $du = (4t - 3) dt$. Thus, $I = \int \frac{du}{u} = \ln |u| + c = \ln |2t^2 - 3t + 4| + c$. Substituting back $t = \sin x$, we get $f(x) = \ln |2 \sin^2 x - 3 \sin x + 4|$. Now, calculate $f\left(\frac{\pi}{2}\right) - f(0)$: $f\left(\frac{\pi}{2}\right) = \ln |2(1)^2 - 3(1) + 4| = \ln |2 - 3 + 4| = \ln 3$. $f(0) = \ln |2(0)^2 - 3(0) + 4| = \ln 4$. Therefore, $f\left(\frac{\pi}{2}\right) - f(0) = \ln 3 - \ln 4 = \ln \left(\frac{3}{4}\right) = \log \left(\frac{3}{4}\right)$.

If $\int \frac{2 \sin 2x - 3 \cos x}{2 \sin^2 x - 3 \sin x + 4} dx = f(x) + c$ where $c$ is the constant of integration, then $f\left(\frac{\pi}{2}\right) - f(0) =$

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