ज्ञात करें `int tan^-1sqrt((1-sinx)/(1+sin x))dx `

Explanation

`(1-sinx)/(1+sinx)=(cos^2""x/2+sin^2""x/2-2cos""x/2sin""x/2)/(cos^2""x/2+sin^2""x/2+2cos""x/2sin""x/2)=((cos""x/2-sin""x/2)/(cos""x/2+sin""x/2))^2`
`:." "sqrt((1-sinx)/(1+sinx))=(cos""x/2-sin""x/2)/(cos""x/2+sin""x/2)=(1-tan""x/2)/(1+tan""x/2)`
`tan(pi/4-x/2)`
अब `I = int tan^(-1){tan(pi/4-x/2)}dx`
`=int(pi/4-x/2)dx=pi/4x-x^2/4+c" "[because" "tan^(-1)(tanx)=x]`