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प्रश्न

\[\int\limits_0^{\pi/2} \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} dx\]
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उत्तर

\[Let\ I = \int_0^\frac{\pi}{2} \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} d\ x . Then, \]
\[\text{Dividing the numerator and denominator by} \cos^2 x, we\ get\]
\[I = \int_0^\frac{\pi}{2} \frac{\sec^2 x}{a^2 \tan^2 x + b^2} d x\]
\[Let\ \tan x = t . Then, \sec^2 x\ dx\ = dt\]
\[When\ x = 0, t = 0\ and\ x\ = \frac{\pi}{2} , t = \infty \]
\[ \therefore I = \int_0^\infty \frac{1}{a^2 t^2 + b^2} d t\]
\[ \Rightarrow I = \frac{1}{a^2} \int_0^\infty \frac{1}{t^2 + \frac{b^2}{a^2}} dt\]
\[ \Rightarrow I = \frac{1}{a^2} \times \frac{a}{b} \left[ \tan^{- 1} \frac{at}{b} \right]_0^\infty \]
\[ \Rightarrow I = \frac{1}{ab}\frac{\pi}{2}\]
\[ \Rightarrow I = \frac{\pi}{2ab}\]

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अध्याय 19: Definite Integrals - Exercise 20.2 [पृष्ठ ३९]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.2 | Q 28 | पृष्ठ ३९

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