∫ln(1+cosx)−xtanx2dx is equal to

∫log1+cosx dx−∫xtanx2dxI1 – I2Integrate I1 by applying integration by parts.∫log(1+cosx).1 dx-∫xtanx2dx =log(1+cosx)(x)-∫11+cosx(-sinx)(x) dx-∫xtanx2dx =xlog(1+cosx)+∫xtanx2dx-∫xtanx2dx+C =xlog(1+cosx)+C

∫ln(1+cosx)−xtanx2dx is equal to

∫log1+cosx dx−∫xtanx2dxI1 – I2Integrate I1 by applying integration by parts.∫log(1+cosx).1 dx-∫xtanx2dx =log(1+cosx)(x)-∫11+cosx(-sinx)(x) dx-∫xtanx2dx =xlog(1+cosx)+∫xtanx2dx-∫xtanx2dx+C =xlog(1+cosx)+C

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$\int (\ln (1 + \cos x) - x \tan \frac{x}{2}) d x i s e q u a l t o$

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$x \ln (1 + \cos x) + C$

$x \ln (1 + \sec x) + C$

$x^{2} \ln (1 + \cos x) + C$

$x lntan x + C$

answer is A.

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Detailed Solution

$\int \log (1 + \cos x) d x - \int x \tan \frac{x}{2} d x$

I₁ – I₂
Integrate I₁ by applying integration by parts.

$\int \log (1 + \cos x) . 1 d x - \int x \tan \frac{x}{2} d x = \log (1 + \cos x) (x) - \int \frac{1}{1 + \cos x} (- \sin x) (x) d x - \int x \tan \frac{x}{2} d x = x \log (1 + \cos x) + \int x \tan \frac{x}{2} d x - \int x \tan \frac{x}{2} d x + C = x \log (1 + \cos x) + C$