ज्ञात करें `int (dx)/(a^2 sin^2 x +b^2 cos^2 x)`

Explanation

`I = int(sec^2x)/(a^2 tan^2 x+b^2)dx = 1/b^2 int (sec^2 x)/(1+a^2/b^2tan^2x)`
`z=a/btanx,` रखे पर, `dz = a/b""sec^(2) xdx`.
अब, `I =1/b^2 ""b/a""int (dz)/(1+z^2)=1/(ab) z+c = 1/(ab)tan^(-1) (a/b tanx)+c`