an interesting integral

[@aviralaryal7332]

Nice video!

[@AliTea1]

amazing did not know you could think of it in this way .

[@holyshit922]

1. Method Rewrite integrand as (1-sin(x))/sqrt(1-sin(x)) and integrate -sin(x)/sqrt(1-sin(x)) by parts with u = 1/sqrt(1-sin(x)) and dv = -sin(x) then symplify using pythagorean trigonometric identity then combine integrals and finally indefinite integral is 2cos(x)/sqrt(1-sin(x)) 2. Method Use substitution u = Pi/2-x then use half angle formula for sine

[@seegeeaye]

@@holyshit922 nice options

[@tgx3529]

I tried use integrál on (0;π/2)+ integrál on (π/2;π), then use sgrt(1-sinx)*sqrt(1+sinx)/sqrt(1+sinx)= |cosx|/sqrt(1+sinx), then substitution sqrt(1+sinx)=y, Ther IT Will be cancel, you Will get 2*integrál from 2 on( 1, sqrt2)=4( sqrt2-1), notes sqrt(1+sinx) Is Always Positive on origin interval.