int left[ log (1+cos x) - x tan left( frac{x} | vedclass

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(D) Let $I = \int \left[ \log (1+\cos x) - x 	an \left( \frac{x}{2} ight) ight] dx$.Using integration by parts on the first term,$\int u \cdot v

Vedclass · Accepted Answer

$x \log |1+\cos x| + c$

Vedclass · Answer

(D) Let $I = \int \left[ \log (1+\cos x) - x 	an \left( \frac{x}{2} ight) ight] dx$.Using integration by parts on the first term,$\int u \cdot v dx = u \int v dx - \int (u' \int v dx) dx$,where $u = \log(1+\cos x)$ and $v = 1$.$I = x \log(1+\cos x) - \int x \cdot \frac{d}{dx} [\log(1+\cos x)] dx - \int x 	an \left( \frac{x}{2} ight) dx$.Since $\frac{d}{dx} [\log(1+\cos x)] = \frac{-\sin x}{1+\cos x} = \frac{-2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)} = -	an(x/2)$.$I = x \log(1+\cos x) - \int x \left( -	an \frac{x}{2} ight) dx - \int x 	an \frac{x}{2} dx$.$I = x \log(1+\cos x) + \int x 	an \frac{x}{2} dx - \int x 	an \frac{x}{2} dx$.$I = x \log(1+\cos x) + c$.

$\int \left[ \log (1+\cos x) - x \tan \left( \frac{x}{2} \right) \right] dx =$

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