int frac{2 tan (x)}{1+2 tan ^2(x)} d x= | vedclass

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(B) We have $I = \int \frac{2 	an x}{1+2 	an^2 x} dx$. Converting to $\sin x$ and $\cos x$: $I = \int \frac{2 \frac{\sin x}{\cos x}}{1+2 \frac{\sin^

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$\log \left|\frac{\cos ^2 x}{2}+\sin ^2 x\right|+c$

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(B) We have $I = \int \frac{2 	an x}{1+2 	an^2 x} dx$. Converting to $\sin x$ and $\cos x$: $I = \int \frac{2 \frac{\sin x}{\cos x}}{1+2 \frac{\sin^2 x}{\cos^2 x}} dx = \int \frac{2 \sin x \cos x}{\cos^2 x + 2 \sin^2 x} dx$. Using $\cos^2 x = 1 - \sin^2 x$,the denominator becomes $(1 - \sin^2 x) + 2 \sin^2 x = 1 + \sin^2 x$. So,$I = \int \frac{2 \sin x \cos x}{1 + \sin^2 x} dx$. Let $t = 1 + \sin^2 x$. Then $dt = 2 \sin x \cos x dx$. Substituting these into the integral: $I = \int \frac{dt}{t} = \log |t| + c_1 = \log |1 + \sin^2 x| + c_1$. Since $1 = \cos^2 x + \sin^2 x$,we have $1 + \sin^2 x = \cos^2 x + 2 \sin^2 x$. $I = \log |\cos^2 x + 2 \sin^2 x| + c_1 = \log |2(\frac{\cos^2 x}{2} + \sin^2 x)| + c_1$. Using $\log(ab) = \log a + \log b$: $I = \log 2 + \log |\frac{\cos^2 x}{2} + \sin^2 x| + c_1 = \log |\frac{\cos^2 x}{2} + \sin^2 x| + c$,where $c = c_1 + \log 2$.

$\int \frac{2 \tan (x)}{1+2 \tan ^2(x)} d x=$

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