Question-122936

Question Number 122936 by CanovasCamiseros last updated on 21/Nov/20

Commented by Dwaipayan Shikari last updated on 21/Nov/20

\int_{0}^{\frac{π}{2}} \frac{d x}{b^{2} {t a n}^{2} x + 1} b t a n x = t \Rightarrow {b s e c}^{2} x = \frac{d t}{d x}

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

= \frac{b}{b^{2} - 1} \int_{0}^{\infty} \frac{1}{t^{2} + 1} - \frac{1}{t^{2} + b^{2}}

= \frac{π b}{2 (b^{2} - 1)} - \frac{π}{2 (b^{2} - 1)}

= \frac{π}{2 (b + 1)}

Commented by CanovasCamiseros last updated on 21/Nov/20

p l e a s e h e l p

Answered by mathmax by abdo last updated on 21/Nov/20

A = \int_{0}^{\frac{π}{2}} \frac{dx}{b^{2} \tan^{2} x + 1} \Rightarrow A =_{tanx = t} \int_{0}^{\infty} \frac{dt}{(1 + t^{2}) (b^{2} t^{2} + 1)}

=_{bt = u} \int_{0}^{\infty} \frac{du}{b (1 + \frac{u^{2}}{b^{2}}) (u^{2} + 1)} = \int_{0}^{\infty} \frac{bdu}{(u^{2} + b^{2}) (u^{2} + 1)}

= \frac{b}{2} \int_{- \infty}^{+ \infty} \frac{du}{(u^{2} + b^{2}) (u^{2} + 1)} (we rake b > 0) let

φ (z) = \frac{1}{(z^{2} + b^{2}) (z^{2} + 1)} \Rightarrow φ (z) = \frac{1}{(z - ib) (z + ib) (z - i) (z + i)}

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, i) + Res (φ, ib)}

Res (φ, i) = \frac{1}{2 i (b^{2} - 1)}

Res (φ, ib) = \frac{1}{2 ib (1 - b^{2})} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {\frac{1}{2 i (b^{2} - 1)} + \frac{1}{2 ib (1 - b^{2})}}

= \frac{π}{b^{2} - 1} - \frac{π}{b (b^{2} - 1)} = \frac{π}{b^{2} - 1} (1 - \frac{1}{b}) = \frac{π (b - 1)}{b (b - 1) (b + 1)} = \frac{π}{b (b + 1)} \Rightarrow

A = \frac{b}{2} . \frac{π}{b (b + 1)} = \frac{π}{2 (b + 1)}