I=∫0πln(1+cosθ)dθ,θ→π−θ I=∫0πln(1−cosθ)dθ 2I=∫0πln(1+cosθ)(1−cosθ)dθ=∫0πlnsin2θdθ =2∫0πlnsinθdθ=4∫0π2lnsinθdθ =4(−π2ln2)=−2πln2→I=−πln2

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$\int_{0}^{π} \ln (1 + \cos θ) d θ$

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$\frac{π}{2} \ln 2$

$π \ln 2$

$- \frac{π}{2} \ln 2$

$- π \ln 2$

answer is D.

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Detailed Solution

$I = \int_{0}^{π} \ln (1 + \cos θ) d θ, θ \to π - θ$

$I = \int_{0}^{π} \ln (1 - \cos θ) d θ$

$2 I = \int_{0}^{π} \ln (1 + \cos θ) (1 - \cos θ) d θ = \int_{0}^{π} \ln \sin^{2} θ d θ$

$= 2 \int_{0}^{π} \ln \sin θ d θ = 4 \int_{0}^{\frac{π}{2}} \ln \sin θ d θ$

$= 4 (- \frac{π}{2} \ln 2) = - 2 π \ln 2 \to I = - π \ln 2$

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