ज्ञात करें `int (dx)/sqrt(a^2-x^2)`

Explanation

`x = a sin theta,` तो `dx=a costheta dtheta`
अब, `int (dx)/(sqrt(a^2-x^2))=int(acos theta)/sqrt(a^2-a^2sin^2theta)d theta=int (acostheta)/sqrt(a^2(1-sin^2theta))d theta`
`=int (acos theta)/(a cos theta)d theta=int d theta `
`=theta +c=sin^(-1)""x/a+c" "[becausesin theta=x/a :. theta = sin^(-1)a/a]`