Question-163288

Question Number 163288 by HongKing last updated on 05/Jan/22

Answered by Ar Brandon last updated on 05/Jan/22

I = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + a^{2} \tan^{2} x} = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{\sec^{2} x + a^{2} \tan^{2} x \sec^{2} x} d x

= \int_{0}^{\infty} \frac{d t}{1 + t^{2} + a^{2} t^{2} (1 + t^{2})} = \int_{0}^{\infty} \frac{d t}{(t^{2} + 1) (a^{2} t^{2} + 1)}

= \int_{0}^{\infty} (\frac{1}{1 - a^{2}} \cdot \frac{1}{t^{2} + 1} + \frac{a^{2}}{a^{2} - 1} \cdot \frac{1}{a^{2} t^{2} + 1}) d t

= \frac{1}{1 - a^{2}} \cdot \frac{π}{2} + \frac{a^{2}}{a^{2} - 1} \cdot \frac{1}{a} \cdot \frac{π}{2} = \frac{1 - a}{1 - a^{2}} \cdot \frac{π}{2} = \frac{π}{2 (1 + a)}

Commented by Ar Brandon last updated on 05/Jan/22

\frac{1}{(t^{2} + 1) (a^{2} t^{2} + 1)} = \frac{p t + q}{t^{2} + 1} + \frac{r t + s}{a^{2} t^{2} + 1}

= \frac{(p t + q) (a^{2} t^{2} + 1) + (r t + s) (t^{2} + 1)}{}

\lim_{t \to i} = (p i + q) (1 - a^{2}) = 1, p = 0, q = \frac{1}{1 - a^{2}}

q + s = 1 \Rightarrow s = 1 - q = \frac{a^{2}}{a^{2} - 1}

p + r = 0 \Rightarrow r = 0

Commented by HongKing last updated on 05/Jan/22

perfect solution my dear Sir thank you

Commented by peter frank last updated on 06/Jan/22

thank you