Evaluate-0-pi-x-a-2-cos-2-x-b-2-sin-2-x-dx-

Question Number 177296 by peter frank last updated on 03/Oct/22

Evaluate

\int_{0}^{π} \frac{x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} dx

Commented by peter frank last updated on 03/Oct/22

ans \frac{π^{2}}{2 ab}

Answered by Ar Brandon last updated on 03/Oct/22

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

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

= π \int_{0}^{\frac{π}{2}} \frac{d (\tan x)}{a^{2} + b^{2} \tan^{2} x} = \frac{π}{a b} {[\arctan (\frac{b \tan x}{a})]}_{0}^{\frac{π}{2}} = \frac{π^{2}}{2 a b}

Commented by peter frank last updated on 03/Oct/22

thank you

Answered by BaliramKumar last updated on 03/Oct/22

I = \int_{0}^{π} \frac{x}{a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x} d x \dots \dots \dots \dots \dots \dots [i]

I = \int_{0}^{π} \frac{(π - x)}{a^{2} {c o s}^{2} (π - x) + b^{2} {s i n}^{2} (π - x)} d x

I = \int_{0}^{π} \frac{π - x}{a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x} d x

I = \int_{0}^{π} \frac{π}{a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x} d x - \int_{0}^{π} \frac{x}{a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x} d x

I = \int_{0}^{π} \frac{π}{a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x} d x - I

2 I = π \int_{0}^{π} \frac{1}{a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x} d x

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

I = \frac{π}{2} \int_{0}^{π} \frac{{s e c}^{2} (π - x)}{a^{2} + b^{2} {t a n}^{2} (π - x)} d x

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

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

l e t b \cdot t a n (x) = y w h e n x = 0 t h e n y = 0

{s e c}^{2} x \cdot d x = \frac{d y}{b} w h e n x = \frac{π}{2} t h e n y = \infty

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

I = \frac{π}{b} \cdot \frac{1}{a} {[{t a n}^{- 1} (\frac{y}{a})]}_{0}^{\infty}

I = \frac{π}{a b} [\frac{π}{2} - 0]

I = \frac{π^{2}}{2 a b}

Commented by peter frank last updated on 03/Oct/22

thank you