By Using properties of definite integrals, evaluate the following…

Question

By Using properties of definite integrals, evaluate the following integral in Exercise:
$\int^{\pi}_{0}\log(1+\cos\text{x})\text{dx}$

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Answer

$\text{Let}\ \text{I}\int^{\pi}\limits_{0}\log(1+\cos\text{x})\text{dx}$$\Rightarrow\ \ \text{I}=\int^{\pi}\limits_{0}\log(1+\cos(\pi-\text{x}))\text{dx}=\int^{\pi}\limits_{0}\log(1-\cos\text{x})\text{dx}$
Adding eq. (i) and (ii)$21=\int^{\pi}\limits_{0}\big[\log(1+\cos\text{x})+\log(1-\cos\text{x})\big]\text{dx}=\int^{\text{x}}\limits_{0}\big[\log(1+\cos\text{x})(1-\cos\text{x})\big]\text{dx}$
$\Rightarrow21=\int^{\pi}\limits_{0}\big[\log(1-\cos^{2}\text{x})\big]\text{dx}=\int^{\pi}\limits_{0}\big[\log\sin^{2}\text{x}\big]\text{dx}=2\int^{\pi}\limits_{0}\big[\log\sin\text{x}\big]\text{dx}$
$\Rightarrow\ \ \text{I}=2\int^{\frac{\pi}{2}}\limits_{0}\log\sin\text{x}\ \text{dx}$
$\Rightarrow\ \ \text{I}=2\int^{\frac{\pi}{2}}\limits_{0}\log\sin\bigg(\frac{\pi}{2}-\text{x}\bigg)\text{dx}=2\int^{\frac{\pi}{2}}\limits_{0}\log\cos\text{x}\ \text{dx}$
Adding eq. (i) and (ii),
$21=2\int^{\frac{\pi}{2}}\limits_{0}(\log\sin\text{x}+\log\cos\text{x})\text{dx}=2\int^{\frac{\pi}{2}}\limits_{0}(\log\sin\text{x}\cos\text{x})\text{dx}$
$\Rightarrow\ \ \text{I}=\int^{\frac{\pi}{2}}\limits_{0}\log\bigg(\frac{2\sin\text{x}\cos\text{x}}{2}\bigg)\text{dx}=\int^{\frac{\pi}{2}}\limits_{0}\log\bigg(\frac{\sin2\text{x}}{2}\bigg)\text{dx}$
$=\int^{\frac{\pi}{2}}\limits_{0}(\log\sin2\text{x}-\log2)\text{dx}$
$$$\Rightarrow\int\limits_{0}^{\frac{\pi}{2}}\log\sin2\text{x}\ \text{dx}-\int\limits_{0}^{\frac{\pi}{2}}\log2\ \text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\log\sin2\text{dx}-\log2\text{(x)}^{\frac{\pi}{2}}_{0}=\int\limits_{0}^{\frac{\pi}{2}}\log\sin2\text{dx}-\frac{\pi}{2}\log2$
$\Rightarrow\ \ \text{I}=\text{I}_{1}-\frac{\pi}{2}\log2$
Where $\text{I}_{1}=\int^{\frac{\pi}{2}}\limits_{0}\log\sin2\text{x}\ \text{dx}$
$\text{putting}\ 2\text{x}=\text{t}\ \text{in}\ \text{eq}.\text{(vi)},\ \Rightarrow2=\frac{\text{dt}}{\text{dx}}\ \Rightarrow\ \ \text{dx}=\frac{\text{dt}}{2}$
Limits of integration when $\text{x}=0, \text{and}\ \text{x}=\frac{\pi}{2},\text{t}=\pi$
$\therefore\ \ \text{from eq. (vi)},\ \ \text{I} _{1}\int^{\pi}\limits_{0}\log\sin\text{t}\frac{\text{dt}}{2}=\frac{1}{2}\int^{\pi}\limits_{0}\log\sin\text{t}\ \text{dt} =\frac{1}{2}\times2\int^{\frac{\pi}{2}}\limits_{0}\log\sin\text{t}\ \text{dt}$
$\Rightarrow\ \ \text{I}_{1}=\int^{\frac{\pi}{2}}\limits_{0}\log\sin\text{t}\ \text{dt}=\int^{\frac{\pi}{2}}\limits_{0}\log\sin\text{x}\text{dx}\ \ \bigg[\because\int^{\text{b}}\limits_{\text{a}}\text{f}\text{(t)}\text{dt}=\int^{\text{b}}\limits_\text{a}\text{f}\text{(x)}\text{dx}\bigg]$
$\Rightarrow\ \ \text{I}=\frac{1}{2}\ \ \ \text{[from eq. (iii)]}$
$\text{putting this value in eq. (v)},\ \text{I}=\frac{1}{2}-\frac{\pi}{2}\log2\ \ \Rightarrow 21=\text{I}-\pi\log2$
$\Rightarrow\ \ \ 21-\text{I}=-\pi\log2\ \ \text{I}=-\pi\log2$

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