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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals4 Marks
Question
By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi} \log (1+\cos x) d x$
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Answer

Given, $\int_{0}^{\pi} \log (1+\cos x) d x$ 
Let, $I=\int_{0}^{\pi} \log (1+\cos x) d x$ .....(i)
as, $\left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\}$ 
$\Rightarrow \mathrm{I}=\int_{0}^{\pi} \log (1+\cos (\pi-\mathrm{x}) \mathrm{d} \mathrm{x}$ 
$\Rightarrow I=\int_{0}^{\pi} \log (1-\cos x) d x$ ......(ii)
Adding (i) and (ii), we get
$2 \mathrm{I}=\int_{0}^{\pi}\{\log (1+\cos \mathrm{x})+\log (1-\cos \mathrm{x})\} \mathrm{d} \mathrm{x}$ 
$2 \mathrm{I}=\int_{0}^{\pi} \log \left(1-\cos ^{2} \mathrm{x}\right) \mathrm{d} \mathrm{x}$ 
$2 \mathrm{I}=\int_{0}^{\pi} \log \left(\sin ^{2} \mathrm{x}\right) \mathrm{d} \mathrm{x}$ 
$2 \mathrm{I}=\int_{0}^{\pi} 2 \cdot \log (\sin \mathrm{x}) \mathrm{d} \mathrm{x}$ 
$2 \mathrm{I}=2 \cdot \int_{0}^{\pi} \log (\sin \mathrm{x}) \mathrm{d} \mathrm{x}$ 
$I=\int_{0}^{\pi} \log (\sin x) d x$ ......(iii)
because, $\int_{0}^{2 a} f(x) d x=2 \cdot \int_{0}^{a} f(x) d x$ if f(2a - x) = f(x)
Here, if f(x) = log (sin x) and f($\pi$ - x) = log ( sin ($\pi$ - x))= log (sin x) = f(x)
$\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}} \log \sin \mathrm{xdx}$ ......(iv)
$\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2}-\mathrm{x}\right) \mathrm{d} \mathrm{x}$ 
$\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}} \log \cos \mathrm{x} \mathrm{d} \mathrm{x}$  ......(v)
Adding (1) and (2), we get
$\Rightarrow 2 \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}}(\log \sin \mathrm{x}+\log \cos \mathrm{x}) \mathrm{d} \mathrm{x}$ 
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}(\log \sin \mathrm{x}+\log \cos \mathrm{x}+\log 2-\log 2) \mathrm{d} \mathrm{x}$ 
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}(\log (2 \sin \mathrm{x} \cos \mathrm{x})-\log 2) \mathrm{d} \mathrm{x}$ 
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}\left(\log (\sin 2 \mathrm{x}) \mathrm{d} \mathrm{x}-\int_{0}^{\frac{\pi}{2}} \log 2 \mathrm{d} \mathrm{x}\right.$
Let 2x = t $\Rightarrow$ 2dx = dt
When x = 0, t = 0 and when x = $\frac{\pi}{4}$, t = $\pi$ 
$\Rightarrow \mathrm{I}=\left[\frac{1}{2} \int_{0}^{\pi}(\log (\sin t) d t]-\left(\frac{\pi}{2} \log 2\right)\right.$ 
$\Rightarrow I=\left[\frac{I}{2}\right]-\left(\frac{\pi}{2} \log 2\right)$ 
$\Rightarrow \mathrm{I}=-\left(\frac{\pi}{2} \log 2\right)$ 
$\Rightarrow \mathrm{I}=-(\pi \log 2)$

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