सिद्ध कीजिए की
`int_(0)^(pi) (x dx)/(1+ sin^(2) x) = pi^(2)/(2sqrt(2))`.

Explanation

माना `I=int_(0)^(pi) (x dx)/(1+ sin^(2) x)` ...(i)
`I=int_(0)^(pi) ((pi-x) dx)/(1+sin^(2) (pi-x))`
`I=int_(0)^(pi) ((pi-x) dx)/(1+sin^(2) x)` ...(ii)
समीकरण (i) तथा (ii) को जोड़ने पर
`2I=int_(0)^(pi) ((x+pi-x)dx)/(1+sin^(2) x) =pi int_(0)^(pi) (dx)/(1+ sin^(2) x)`
`=2 pi int_(0)^(pi//2) (dx)/(1+sin^(2) x)` [प्रगुण (6) से]
`=2pi int_(0)^(pi//2) (cosec^(2) x dx)/(cosec^(2) x+1)`
`=2pi int_(0)^(pi//2) (cosec^(2) x dx)/(2+ cot^(2) x)` माना `cot=t, cpsec^(2) x dx =dt`
`=2pi int_(0)^(pi//2) (dt)/(2+t^(2))`
`=2pi int_(0)^(oo) (dt)/((sqrt(2))^(2)+t^(2))`
`=2pi [1/sqrt(2) "tan"^(-1) 1/sqrt(2)]_(0)^(oo)=pisqrt(2) [tan^(-1) oo-tan^(-1) 0]`
`=pisqrt(2) (pi/2)=pi^(2)/sqrt(2)`
`I=pi^(2)/(2sqrt(2))`