0-pi-2-2xsin-2-t-cos-2-t-x-2-sin-2-t-dt-

Question Number 143962 by lapache last updated on 20/Jun/21

\int_{0}^{\frac{π}{2}} \frac{2 {x s i n}^{2} t}{{c o s}^{2} t + x^{2} {s i n}^{2} t} d t = \dots . . ? ? ?

Answered by Dwaipayan Shikari last updated on 20/Jun/21

\frac{2}{x} \int_{0}^{\frac{π}{2}} \frac{x^{2} {s i n}^{2} t}{{c o s}^{2} t + x^{2} {s i n}^{2} t} d t

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

= \frac{π}{x} - \int_{0}^{\infty} \frac{1}{(1 + {(x u)}^{2}) (1 + u^{2})} d u

= \frac{π}{x} + \frac{1}{x^{2} - 1} \int_{0}^{\infty} \frac{1}{1 + x^{2} u^{2}} - \frac{1}{1 + u^{2}} d u

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

Commented by mnjuly1970 last updated on 20/Jun/21

p l e a s e r e c h e c h p e n - u l t i m a t e l i n e

(c o e f f i c i e n t)

Answered by ArielVyny last updated on 20/Jun/21

\int_{0}^{\frac{π}{2}} \frac{1}{\frac{{c o s}^{2} t + x^{2} {s i n}^{2} t}{2 {x s i n}^{2} t}} d t

\int_{0}^{\frac{π}{2}} \frac{1}{\frac{1}{2 x} \times \frac{1}{{t a n}^{2} t} + \frac{x}{2}} d t = 2 \int_{0}^{\frac{π}{2}} \frac{1}{\frac{1}{{x t a n}^{2} t} + x} d t

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