ज्ञात करें `int (dx)/(sqrt(a^2+x^2))`

Explanation

यहां समाकलय में `sqrt(a^2 + x^2)` के रूप का व्यंजनक है, इसलिए `x=a tan theta` रखे ]
माना की `x = a tan ` तो `dx = a sec^2 theta d theta`
अब, `int (dx)/(sqrt(a^2+x^2))=int (asec^2 theta)/sqrt(a^2+a^2 tan^2 theta)d theta=int (a sec^2 theta)/sqrt(a^2(1+tan^2theta))d theta`
`int (a sec^2 theta)/(a sec theta)d theta= int sec theta d theta`
`= log | sec theta+tan theta| +c " "[because int sec theta d theta = log| sec theta + tan theta|]`
`because x = a tan theta " ":." "tan theta = x/a `
` :." "sec theta = sqrt( 1 + tan^(2) theta) = sqrt(1+x^2/a^2)=(sqrt(a^2 +x^2))/a`
अब (1) से, int `(dx)/(sqrt(a^2 +x^2))= log | (sqrt( a^2+x^2))/a+x/a|+c`
`log (sqrt(a^2+x))-log |a|+c" "[ because x + sqrt(a^2+x^2), theta]`
` = log (x+ sqrt(a^2 +x^2))+k,` जहां `k= c - log |a|`+N74