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UPSC NDA Math Model Paper 9

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Question : 3 of 120

Marks: +1, -0

What is

\int\limits_{0}^{\frac{\pi}{2}} \frac{dx}{a^{2} \cos^{2} x + b^{2} \sin^{2} x}

2ab
2πab
$\frac{\pi}{2ab}$
$\frac{\pi}{ab}$

Solution:

Let,

I = \int\limits_{0}^{\frac{\pi}{2}} \frac{dx}{a^{2} \cos^{2} x + b^{2} \sin^{2} x}

I = \int\limits_{0}^{\frac{\pi}{2}} \frac{dx}{\cos^{2} x \left( a^{2} + b^{2} \frac{\sin^{2} x}{\cos^{2} x} \right)}

= \int\limits_{0}^{\frac{\pi}{2}} \frac{\sec^{2} x \, dx}{a^{2} + b^{2} \tan^{2} x}

Let

\tan x = t

\Rightarrow \sec^{2} x \, dx = dt

When x = 0, t = tan (0) = 0

When x = π/2, t = ∞

I = \int\limits_{0}^{\infty} \frac{dt}{a^{2} + b^{2} t^{2}}

= \frac{1}{b^{2}} \int\limits_{0}^{\infty} \frac{dt}{\left( \left( \frac{a}{b} \right)^{2} + t^{2} \right)}

= \frac{1}{b^{2}} \left[ \frac{b}{a} \tan^{-1} \left( \frac{b t}{a} \right) \right]_{0}^{\infty}

= \frac{1}{a b} \left[ \tan^{-1}(\infty) - \tan^{-1}(0) \right]

= \frac{1}{a b} \times \left( \frac{\pi}{2} - 0 \right)

= \frac{\pi}{2 a b}

Hence, option (3) is correct.

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