Integrate (sinx+cosx)/√(sin2x)?

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Answer 1 · 2017-09-10T17:27:15

$a r c sin (sin x - cos x) + C, or, a r c sin {\sqrt{2} sin (x - \frac{π}{4})} + C .$ $arc sin(sinx-cosx)+C, or, arc sin{sqrt2sin(x-pi/4)}+C.$

Let, $I = \int \frac{sin x + cos x}{\sqrt{sin 2 x}} d x .$ $I=int(sinx+cosx)/sqrt(sin2x)dx.$

We subst., $sin x - cos x = t \Rightarrow (cos x + sin x) d x = d t .$ $sinx-cosx=t rArr (cosx+sinx)dx=dt.$

Further,

$t^{2} = {(sin x - cos x)}^{2} = {sin}^{2} x - 2 sin x cos x + {cos}^{2} x, i . e .,$ $t^2=(sinx-cosx)^2=sin^2x-2sinxcosx+cos^2x, i.e.,$

$t^{2} = 1 - sin 2 x \Rightarrow sin 2 x = 1 - t^{2} .$ $t^2=1-sin2x rArr sin2x=1-t^2.$

$:. I=intdt/sqrt(1-t^2),$

$=arc sint,$

$rArr I=arc sin(sinx-cosx)+C.$

Since, $sinx-cosx=sqrt2(1/sqrt2sinx-1/sqrt2cosx),$

$=sqrt2{sinxcos(pi/4)-cosxsin(pi/4)},$

$=sqrt2sin(x-pi/4),$ we can write,

$I=arc sin{sqrt2sin(x-pi/4)}+C.$

Enjoy Maths!