First slide

Evaluation of definite integrals

Question

$\int_{0}^{π} \log (1 + \cos x) dx$ is equal to

Moderate

A

0
B

$\frac{π}{2} \log 2$
C

$- πlog 2$
D

$2 πlog 2$

Solution

Let,
$\begin{matrix} I = \int_{0}^{π} \log (1 + \cos x) dx - - - - (i) \\ I = \int_{0}^{π} \log {1 + \cos (π - x)} dx \\ = \int_{0}^{π} \log (1 - \cos x) dx [∵ \cos (π - x) = - \cos x] \dots (ii) \\ = \int_{0}^{π} \log \{2 \sin^{2} (\frac{x}{2})\} dx [∵ 1 - \cos x = 2 \sin^{2} \frac{x}{2}] \\ = \int_{0}^{π} \{\log 2 + 2 \log (\sin \frac{x}{2})\} dx \\ = \int_{0}^{π} \log 2 dx + 2 \int_{0}^{π} \log (\sin \frac{x}{2}) dx \end{matrix}$
In the second. integral, put $\frac{x}{2} = t \Rightarrow dx = 2 dt$
and limits when x =0, t =0 and when $x = π, t = π / 2$
$\begin{matrix} ∴ I = \log 2 [x]_{0}^{π} + 2 \int_{0}^{π / 2} \log (\sin t) 2 dt \\ = (\log 2) (π - 0) + 4 (- \frac{π}{2} \log 2) \\ [∵ \int_{0}^{π / 2} \log \sin xdx = - \frac{π}{2} \log 2] \end{matrix}$
$= - πlog 2$

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