0-pi-xdx-a-2-cos-2-x-b-2-sin-2-x-2-

Question Number 151429 by peter frank last updated on 21/Aug/21

\int_{0}^{π} \frac{xdx}{{(a^{2} \cos^{2} x + b^{2} \sin^{2} x)}^{2}}

Answered by Olaf_Thorendsen last updated on 21/Aug/21

I = \int_{0}^{π} \frac{x}{{(a^{2} \cos^{2} x + b^{2} \sin^{2} x)}^{2}} d x

Let u = π - x :

I = \int_{0}^{π} \frac{π - u}{{(a^{2} \cos^{2} u + b^{2} \sin^{2} u)}^{2}} d u

\Rightarrow 2 I = \int_{0}^{π} \frac{π}{{(a^{2} \cos^{2} u + b^{2} \sin^{2} u)}^{2}} d u

\Rightarrow I = \frac{π}{2} \int_{0}^{π} \frac{d u}{a^{4} \cos^{4} u {(1 + \frac{b^{2}}{a^{2}} \tan^{2} u)}^{2}}

I = π \int_{0}^{\frac{π}{2}} \frac{d u}{a^{4} \cos^{4} u {(1 + \frac{b^{2}}{a^{2}} \tan^{2} u)}^{2}}

Let t = \tan u :

I = π \int_{0}^{\infty} \frac{\frac{d t}{1 + t^{2}}}{a^{4} \frac{1}{{(1 + t^{2})}^{2}} {(1 + \frac{b^{2}}{a^{2}} t^{2})}^{2}}

I = \frac{π}{a^{4}} \int_{0}^{\infty} \frac{1 + t^{2}}{{(1 + \frac{b^{2}}{a^{2}} t^{2})}^{2}} d t

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

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

I = \frac{π}{a^{2} b^{2}} [\frac{a}{b} \arctan (\frac{b}{a} t) + \frac{a^{2} - b^{2}}{2} . \frac{t}{(a^{2} + b^{2} t^{2})}

{+ \frac{a^{2} - b^{2}}{2 a b} \arctan (\frac{b}{a} t)]}_{0}^{\infty}

I = \frac{π}{a^{2} b^{2}} (\frac{π}{2} . \frac{a}{b} + 0 + \frac{π}{2} . \frac{a^{2} - b^{2}}{2 a b})

I = \frac{π^{2} (3 a^{2} - b^{2})}{4 a^{3} b^{3}}