Question

Integral of square root of 1-sin2x

Answer 1

I = √(1 - sin2x ).dx

we know,
sin²x + cos²x = 1
and
sin2x = 2sinx.cosx put it here,

I = √{ sin²x + cos²x -2sinx.cosx } dx

I = √{ (sinx - cosx)²} dx

I = | sinx - cosx |.dx

I = ( sinx.dx - cosx.dx ) { here mod contain if here given Limit then I break mod wth condition of limit . but this is indefinite integration so, we can remove mod without using any sign

I = ( sinx - cosx )dx

I = ( sinx.dx - cosx.dx )

I = ( -cosx - sinx) + C

method 2 :-

we know ,
sin2x = cos(π/2 -2x )
and,
1 - cos(π/2 -2x ) = 2sin²( π/4 - x)

use this ,.

I = √{ 2sin²(π/4 - x) } dx

I = √2 | sin(π/4 - x) | dx

I = -√2 cos( π/4 - x) + C

Answer 2

Hi there

here's your answer

hope it helps