ज्ञात करें `int(dx)/(1-sinx)`

Explanation

`int (dx)/(1-sin x)=int(dx)/(1-cos(pi//2-x))`
`=int(dx)/(2sin^2(pi//4-x//2))" "[because 1 - cos theta=2sin^2""theta/2]`
`1/2int"cosec"^2(pi/4-x/2)dx=1/2([-cot(pi//4-x//2)])/(-1/2)+c=cot(pi/4-x/2)+c`.