Mathematics

# Evaluate: $\displaystyle \int_{0}^{\pi} \dfrac {x}{a^{2} \cos^{2}x + b^{2}\sin^{2}x} dx$

##### SOLUTION
We have, $I = \displaystyle \int_{0}^{\pi} \dfrac {x}{a^{2} \cos^{2}x + b^{2} \sin^{2}x}dx$  .... (i)
$I =\displaystyle \int_{0}^{\pi} \dfrac {\pi - x}{a^{2}\cos^{2} (\pi - x) + b^{2}\sin^{2} (\pi - x)}dx$
$\left (\because \displaystyle \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x)dx\right )$
$I = \displaystyle \int_{0}^{\pi} \dfrac {\pi - x}{a^{2}\cos^{2}x + b^{2}\sin x}dx$  .... (ii)
On adding equations (i) and (ii), we get
$2I = \displaystyle \int_{0}^{\pi} \dfrac {x + \pi - x}{a^{2}\cos^{2} x + b^{2} \sin^{2}x}dx$
$2I = \pi \displaystyle \int_{0}^{\pi} \dfrac {1}{a^{2} \cos^{2}x + b^{2}\sin^{2} x}dx$
Using the property, $\int_{0}^{2a} f(x) \cdot d = \int_{0}^{a} [f(x) + f(2a - x)]dx$
$\therefore 2I = \pi \left [\displaystyle \int_{0}^{\pi /2} \dfrac {1}{a^{2}\cos^{2} x + b^{2} \sin^{2} x} dx + \int_{0}^{\pi/2} \dfrac {1}{a^{2}\cos^{2} (\pi - x) + b^{2}\sin^{2} (\pi - x)}dx\right ]$
$I = \pi \displaystyle \int_{0}^{\pi/2} \dfrac {1}{a^{2}\cos^{2} x + b^{2}\sin^{2} x}dx = \pi \int_{0}^{\pi/2} \dfrac {dx}{a^{2}\cos^{2}x\left (1 + \dfrac {b^{2}}{a^{2}}\tan^{2}x\right )}$
$= \dfrac {\pi}{a^{2}} \displaystyle \int_{0}^{\pi/2} \dfrac {\sec^{2}xdx}{1 + \dfrac {b^{2}}{a^{2}}\tan^{2}x}$ [By putting $u = \dfrac {b}{a}\tan x, du = \dfrac {b}{a}\sec^{2}x dx$ when $x = 0, u = 0$ when $x = \dfrac {\pi}{2}, u = \infty]$
$= \dfrac {\pi}{a^{2}} \displaystyle \int_{0}^{\infty} \dfrac {\dfrac {a}{b}du}{1 + u^{2}}$
$= \dfrac {\pi}{ab} \displaystyle \int_{0}^{\infty} \dfrac {du}{1 + u^{2}} = \dfrac {\pi}{ab} [\tan^{-1} (u)]_{0}^{\infty} + c$
$= \dfrac {\pi}{ab} [\tan^{-1} \infty - \tan^{-1} 0] + c$
$= \dfrac {\pi}{ab} \left [\dfrac {\pi}{2} - 0\right ] + c$
$= \dfrac {\pi^{2}}{2ab}$

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Subjective Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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