ज्ञात करें `int (dx)/(""_((a^2-x^2)^(3/2)))`

Explanation

माना की `x=a sin theta,` तो `dx = acos theta d theta`
अब, `int(dx)/(""_((a^2-x^2)^(3/2)))=int(a cos theta d theta)/(""_((a^2-x^2sin^2 theta)^(3/2)))=int(a cos theta)/(""_(a^2[(1-sin^2 theta)]^(3/2)))`
`=int(a cos theta)/(""_((a^2 cos^2 theta)^(3/2)))d theta = int (a cos theta)/(a^3 cos ^3 theta)d theta`
`=1/a^2int 1/(cos^2theta)d theta = 1/a^2 int sec^2 theta d theta`
`=1/a^2 tan theta+c" ...(1)"`
`because" sin"theta=x/a" ":." "tantheta=x/sqrt(a^2-x^2)`
`:." (1)"` से, `int=(dx)/(""_((a^2-x^2)^(3/2)))=1/a^2x/sqrt(a^2-x^2)+c`