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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_0^x \frac{d x}{1-2 a \cos x+a^2}=$
  • A
    $\frac{\pi}{2\left(1-a^2\right)}$
  • B
    $\pi\left(1-a^2\right)$
  • ✓
    $\frac{\pi}{1-a^2}$
  • D
    $2 \pi\left(1-a^2\right)$
✓

Answer

Correct option: C.
$\frac{\pi}{1-a^2}$
(C)
$\int_0^\pi \frac{ d x}{1-2 a \cos x+ a ^2}$
$=\int_0^\pi \frac{ d x}{\left(1+ a ^2\right)\left(\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}\right)-2 a \left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right)}$
$=\int_0^\pi \frac{ d x}{(1- a )^2 \cos ^2 \frac{x}{2}+(1+ a )^2 \sin ^2 \frac{x}{2}}$
$=\int_0^\pi \frac{\sec ^2 \frac{x}{2}}{(1-a)^2+\left(1+a^2\right) \tan ^2 \frac{x}{2}} d x$
$=\frac{2}{(1+a)^2} \int_0^{\infty} \frac{d t}{\left\{\frac{(1-a)}{(1+a)}\right\}^2+t^2}$ $\ldots \left[\right.$ Put $\left.t =\tan \frac{x}{2} \Rightarrow dt =\frac{1}{2} \sec ^2 \frac{x}{2} d x\right]$
$\begin{array}{l}=\frac{2}{(1+a)^2} \cdot \frac{(1+a)}{(1-a)}\left[\tan ^{-1}\left(\frac{1+a}{1-a} \cdot t\right)\right]_0^{\infty} \\ =\frac{2}{(1+a)^2}\left(\tan ^{-1} \infty-\tan ^{-1} 0\right) \\ =\frac{\pi}{1-a^2}\end{array}$

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