The value of the definite integral ∫0πln1+cosxdx is

I=∫0πln(1+cosx)dx;I=∫0πln(1−cosx)dx 2I=∫0πln(1−cos2x)dx=∫0πln(sin2x)dx 2I=2∫0πln(sinx)dx I=∫0πln(sinx)dx=2∫0π/2ln(sinx)dx=−2(π2)ln2=−πln2

The value of the definite integral ∫0πln1+cosxdx is

I=∫0πln(1+cosx)dx;I=∫0πln(1−cosx)dx 2I=∫0πln(1−cos2x)dx=∫0πln(sin2x)dx 2I=2∫0πln(sinx)dx I=∫0πln(sinx)dx=2∫0π/2ln(sinx)dx=−2(π2)ln2=−πln2

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The value of the definite integral $\int_{0}^{π} \ln (1 + \cos x) d x$ is

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$- π \ln 2$

$\frac{- π}{2} \ln 2$

$- 2 π \ln 2$

$\frac{- π}{4} \ln 2$

answer is A.

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Detailed Solution

$I = \int_{0}^{π} l n (1 + \cos x) d x; I = \int_{0}^{π} l n (1 - \cos x) d x 2 I = \int_{0}^{π} l n (1 - \cos^{2} x) d x = \int_{0}^{π} l n (\sin^{2} x) d x 2 I = 2 \int_{0}^{π} l n (\sin x) d x I = \int_{0}^{π} l n (\sin x) d x = 2 \int_{0}^{π / 2} l n (\sin x) d x = - 2 (\frac{π}{2}) l n 2 = - π l n 2$