ज्ञात करें `int sqrt(sec x -1)dx`

Explanation

`int sqrt(sec x -1)dx = int sqrt((1-cos x)/(cos x))dx`
` =int sqrt((2sin^(2)x/2)/(cosx))dx = sqrt2 int (sin""x/2)/sqrt(2 cos^2""x/2-1)dx`
`z = sqrt2 cos""x/2` रखे, तो `dz = -sqrt2 sin ""x/2 ""1/2 dx`
`:." "sin""x/2 dx = -sqrt2 dz`
अब `I = sqrt2 int (-sqrt2)/sqrt(z^2 -1)dz = -2log|z+sqrt(z^2 -1)|+c`
`=-2 log|sqrt2 cos""x/2 +sqrt(2cos^2 ""x/2 -1)|+c`
`=-2log | sqrt2 cos ""x/2 +sqrt(cos x) | +c`