ज्ञात करें `int (e^(-x/2) sqrt(1-sinx))/(1+cos x)dx`

Explanation

`z =-x/2,` रखे तो `x = -2z` तथा `dx = -2dz`
अब, `I = int ( e^zsqrt(1-sin(-2z)))/(1+cos (-2z))(-2)dz =-2 int (e^z sqrt(1+sin 2pi))/(1+cos 2z)dz`
`=-2 int e^z (cos z+sin z)/(2 cos^2z)dz = -int e^z (sec z+sec z tan z) dz `
`=-int e^z f(z) +c=-e^z sec z +c`
`=-e^(-x/2)sec(-x/2)+c =-e^(-x/2)sec""x/2 +c`