Definite Integration
Parametric Differentiation — Classic
Grade 12
Question:
<p>Evaluate \(I(a)=\displaystyle\int_0^{\pi}\ln(1-2a\cos x+a^2)\,dx\) for \(|a|<1\) and \(|a|>1\). [JEE Advanced 2010]</p>
0 for |a|<1, 2\pi ln|a| for |a|>1
\pi ln|a|
2\pi ln|a| for all a
0 for all a
Step-by-Step Solution
Key Concept: Differentiate under integral: I'(a) = \int_0^\pi (-2cosx+2a)/(1-2acosx+a^2)dx. For |a|<1: I'(a)=0, I(0)=0 \to I=0. For |a|>1: I'(a)=2\pi/a, integrate \to I=2\pi ln|a|.
<div class='solution'>
<p>$I'(a)=\int_0^\pi\frac{2(a-\cos x)}{1-2a\cos x+a^2}dx$.</p>
<p>Using Poisson kernel: $\int_0^\pi\frac{a-\cos x}{1-2a\cos x+a^2}dx = \begin{cases}0 & |a|<1\\\pi/a & |a|>1\end{cases}$</p>
<p>So $I'(a)=\begin{cases}0&|a|<1\\2\pi/a&|a|>1\end{cases}$.</p>
<p>Integrating: $I(a)=\begin{cases}C_1&|a|<1\\2\pi\ln|a|+C_2&|a|>1\end{cases}$</p>
<p>$I(0)=\int_0^\pi\ln 1\,dx=0\Rightarrow C_1=0$. As $a\to 1^+$: $I\to 0\Rightarrow C_2=0$.</p>
<p>$$\boxed{I(a)=\begin{cases}0&|a|<1\\2\pi\ln|a|&|a|>1\end{cases}}$$</p>
</div>
Correct Answer: A