ज्ञात करें ` int (dx)/sqrt(x^2 -a^2)`

Explanation

माना की `x = a sec theta, ` तो `dx = a sec theta tan theta d theta`
अब, ` int (dx)/sqrt(x^2-a^2)= int (a sec theta tan theta)/sqrt(a^2 sec^2 theta - a^2) d theta = int(a sec theta tan theta)/sqrt(a^2 (sec^(2) theta -1)) d theta`
`= int (a sec theta tan theta)/(a tan theta)d theta = int = sec theta d theta `
`log | sec theta + tan theta|+c" ...(1)"`
`because tan theta =sqrt(sec^2 theta - 1)=sqrt(x^2/a^2)-1=sqrt(x^2-a^2)/a`
अतः (1) से, `int (dx)/(sqrt(x^2-a^2))=log |x/a+sqrt(x^2+a^2)/a|+c=log|(x+sqrt(x^2-a^2))/a|+c`
`= log |x+sqrt(x^2-a^2)|-log |a|+c=log|x+sqrt(x^2 -a^2)+k`