If f(x) = int frac{2-3 sin^2 x}{1+cos 2x} dx | vedclass

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(A) Given $f(x) = \int \frac{2-3 \sin^2 x}{1+\cos 2x} dx$. Using the identity $1+\cos 2x = 2 \cos^2 x$, we have: $f(x) = \int \frac{2-3 \sin^2 x}{2 \c

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$\frac{3}{8}(4-\pi)$

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(A) Given $f(x) = \int \frac{2-3 \sin^2 x}{1+\cos 2x} dx$. Using the identity $1+\cos 2x = 2 \cos^2 x$, we have: $f(x) = \int \frac{2-3 \sin^2 x}{2 \cos^2 x} dx = \int \left( \frac{2}{2 \cos^2 x} - \frac{3 \sin^2 x}{2 \cos^2 x} ight) dx$ $f(x) = \int (\sec^2 x - \frac{3}{2} 	an^2 x) dx$ Since $	an^2 x = \sec^2 x - 1$, we get: $f(x) = \int (\sec^2 x - \frac{3}{2}(\sec^2 x - 1)) dx = \int (\sec^2 x - \frac{3}{2} \sec^2 x + \frac{3}{2}) dx$ $f(x) = \int (\frac{3}{2} - \frac{1}{2} \sec^2 x) dx = \frac{3}{2} x - \frac{1}{2} 	an x + C$ Given $f\left(\frac{\pi}{4}ight) = 1$: $\frac{3}{2} \left(\frac{\pi}{4}ight) - \frac{1}{2} 	an\left(\frac{\pi}{4}ight) + C = 1$ $\frac{3\pi}{8} - \frac{1}{2}(1) + C = 1 \Rightarrow C = 1 + \frac{1}{2} - \frac{3\pi}{8} = \frac{3}{2} - \frac{3\pi}{8} = \frac{12-3\pi}{8} = \frac{3}{8}(4-\pi)$ Now, $f(0) = \frac{3}{2}(0) - \frac{1}{2} 	an(0) + C = C = \frac{3}{8}(4-\pi)$.

If $f(x) = \int \frac{2-3 \sin^2 x}{1+\cos 2x} dx$ and $f\left(\frac{\pi}{4}\right) = 1$, then $f(0) =$

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