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Definite Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\limits^{\text{a}}_{-\text{a}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$
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Answer

Let $\text{I}=\int\limits^{\text{a}}_{-\text{a}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$
Consider $\text{f}(\theta)=\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)$
$\text{f}(-\theta)=\log\Big(\frac{\text{a}-\sin(-\theta)}{\text{a}+\sin(-\theta)}\Big)$
$=\log\Big(\frac{\text{a}+\sin\theta}{\text{a}-\sin\theta}\Big)$ $[\sin(-\text{x})=-\sin\text{x}\big]$
$=\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)^{-1}$
$=-\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)$ $\Big[\log\text{a}^{\text{b}}=\text{b}\log\text{a}\Big]$
$=-\text{f}(\theta)$
$\therefore\ \text{f}(-\theta)=-\text{f}(\theta)$
$\Rightarrow\text{I}=\int\limits^{\text{a}}_{-\text{a}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta=0$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$

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