(i) intregation of √1-sin x dx.
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here is the best answer ✔️✔️✔️
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
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