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Model Paper 4 — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 43 Marks
Question
Evaluate the integral: $\int_0^\pi \frac{x}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x$
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Answer

We have,
$I=\int_0^\pi \frac{x}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x$
Using property of definite integral we have,
$=\int_0^\pi \frac{(\pi-x)}{a^2 \cos ^2(\pi-x)+b^2 \sin ^2(\pi-x)} d x$
$=\int_0^\pi \frac{\pi-x}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x \ldots \text { (ii) }$
Adding $(i)$ and $(ii)$
$2 I=\int_0^\pi \frac{x+\pi-x}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x$
$=\pi \int_0^\pi \frac{1}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x$
$=\pi \int_0^\pi \frac{\sec ^2 x}{a^2+b^2 \tan ^2 x} d x \ldots\ ($Dividing numerator and denominator by $\cos ^2 x) $
$=2 \pi \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{a^2+b^2 \tan ^2 x} d x \ldots[$Using $\int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_0^a f(2 a-x) d x]$
Put $\tan x = t$
$\Rightarrow \sec ^2 x d x=d t$
When $x \rightarrow 0 ; t \rightarrow 0$
and $x \rightarrow \frac{\pi}{2} ; t \rightarrow \infty$
$\therefore 2 I=2 \pi \int_0^{\frac{x}{2}} \frac{d t}{a^2+b^2 t^2}$
$\Rightarrow I=\frac{\pi}{b^2} \int_0^{\frac{\pi}{2}} \frac{d t}{\frac{a^2}{b^2}+t^2}$
$=\frac{\pi}{b^2} \times \frac{b}{a}\left[\tan ^{-1}\left(\frac{b t}{a}\right)\right]_0^{\infty}$
$=\frac{\pi}{a b}\left[\frac{\pi}{2}-0\right]$
$=\frac{\pi}{a b} \times \frac{\pi}{2}$
$=\frac{\pi^2}{2 a b}$
Hence, $I=\frac{\pi^2}{2 a b}$

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