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

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(A) दिया गया है $f(x) = \int \frac{2-3 \sin^2 x}{1+\cos 2x} dx$। सर्वसमिका $1+\cos 2x = 2 \cos^2 x$ का उपयोग करने पर: $f(x) = \int \frac{2-3 \sin^2 x}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) दिया गया है $f(x) = \int \frac{2-3 \sin^2 x}{1+\cos 2x} dx$। सर्वसमिका $1+\cos 2x = 2 \cos^2 x$ का उपयोग करने पर: $f(x) = \int \frac{2-3 \sin^2 x}{2 \cos^2 x} dx = \int \left( \frac{2}{2 \cos^2 x} - \frac{3 \sin^2 x}{2 \cos^2 x} ight) dx$ $f(x) = \int (\sec^2 x - \frac{3}{2} 	an^2 x) dx$ चूंकि $	an^2 x = \sec^2 x - 1$, इसलिए: $f(x) = \int (\sec^2 x - \frac{3}{2}(\sec^2 x - 1)) dx = \int (\sec^2 x - \frac{3}{2} \sec^2 x + \frac{3}{2}) dx$ $f(x) = \int (\frac{3}{2} - \frac{1}{2} \sec^2 x) dx = \frac{3}{2} x - \frac{1}{2} 	an x + C$ दिया गया है $f\left(\frac{\pi}{4}ight) = 1$: $\frac{3}{2} \left(\frac{\pi}{4}ight) - \frac{1}{2} 	an\left(\frac{\pi}{4}ight) + C = 1$ $\frac{3\pi}{8} - \frac{1}{2}(1) + C = 1 \Rightarrow C = 1 + \frac{1}{2} - \frac{3\pi}{8} = \frac{3}{2} - \frac{3\pi}{8} = \frac{3}{8}(4-\pi)$ अब, $f(0) = \frac{3}{2}(0) - \frac{1}{2} 	an(0) + C = C = \frac{3}{8}(4-\pi)$।

यदि $f(x) = \int \frac{2-3 \sin^2 x}{1+\cos 2x} dx$ और $f\left(\frac{\pi}{4}\right) = 1$ है, तो $f(0) =$

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