If int cos x log left(tan frac{x}{2}right) dx | vedclass

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(B) Let $I = \int \cos x \log \left(	an \frac{x}{2}ight) dx$. Using integration by parts, $\int u v dx = u \int v dx - \int (u' \int v dx) dx$, whe

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$c-x$

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(B) Let $I = \int \cos x \log \left(	an \frac{x}{2}ight) dx$. Using integration by parts, $\int u v dx = u \int v dx - \int (u' \int v dx) dx$, where $u = \log \left(	an \frac{x}{2}ight)$ and $v = \cos x$. Then $u' = \frac{1}{	an(x/2)} \cdot \sec^2(x/2) \cdot \frac{1}{2} = \frac{1}{2 \sin(x/2) \cos(x/2)} = \frac{1}{\sin x}$ and $\int v dx = \sin x$. $I = \log \left(	an \frac{x}{2}ight) \cdot \sin x - \int \left( \frac{1}{\sin x} \cdot \sin x ight) dx$. $I = \sin x \log \left(	an \frac{x}{2}ight) - \int 1 dx$. $I = \sin x \log \left(	an \frac{x}{2}ight) - x + c$. Comparing this with the given expression $\sin x \log \left(	an \frac{x}{2}ight) + f(x)$, we get $f(x) = c - x$.

If $\int \cos x \log \left(\tan \frac{x}{2}\right) dx = \sin x \log \left(\tan \frac{x}{2}\right) + f(x)$, then $f(x)$ is equal to (assuming $c$ is an arbitrary real constant).

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