1. ## Integrating sqrt(1-sin2x)

Hello,

I'd really appreciate some help with finding the following indefinite integral:

$\int \sqrt{1- \sin 2x}$ $dx$

What I tried is the following:

$\int \sqrt{1- \sin 2x}$ $dx=$

$\int \sqrt{1- 2\sin x \cos x}$ $dx=$

$\int \sqrt{\sin^2x + \cos^2 x - 2\sin x \cos x}$ $dx=$

$\int \sqrt{\sin^2x - 2\sin x \cos x + \cos^2 x }$ $dx=$

$\int \sqrt{(\sin x - \cos x )^2}$ $dx=$

but it would be WRONG to proceed like this:
$\int \sqrt{(\sin x - \cos x )^2}$ $dx=$
$\int (\sin x - \cos x )$ $dx=$
$\int \sin x$ $dx$ $- \int \cos x$ $dx=$
$-\cos x -\sin x +C$

because $\sqrt{a^2}$ could be either $a$ or $-a$ !

So perhaps this problem needs some other approach.

Mathematica gives this:

$\frac {(\cos x + \sin x)\sqrt {1 - \sin 2 x}} {\cos x - \sin x}$

Thanks a lot for your help!

proof

(1-sin2x)^1/2=sinx-cosx your solution substitute it in

$\frac {(\cos x + \sin x)\sqrt {1 - \sin 2 x}} {\cos x - \sin x}
$

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# integration of 1 sin2x

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