integration of (1 - sin2x)^(1/2), sin^2x + cos^2x - 2sinxcosx

[cookie]

What am I doing wrong?

∫[(1-sin2x)^1/2]dx = ∫ [(sin^2x + cos^2x - 2sinxcosx)^1/2] dx= ∫[(cosx - sinx)^2^1/2] dx = sinx + cosx + C

I'm supposed to get sinx - cosx + C

Another question...
sin^2x + cos^2x - 2sinxcosx = (sinx - cosx)^2 or (cosx - sinx)^2?

 

[skeeter]

What am I doing wrong?

∫[(1-sin2x)^1/2]dx = ∫ [(sin^2x + cos^2x - 2sinxcosx)^1/2] dx= ∫[(cosx - sinx)^2^1/2] dx = sinx + cosx + C

I'm supposed to get sinx - cosx + C

Another question...
sin^2x + cos^2x - 2sinxcosx = (sinx - cosx)^2 or (cosx - sinx)^2? Both

I don't agree with

I'm supposed to get sinx - cosx + C

\(\displaystyle \int \sqrt{(\cos{x}-\sin{x})^2} \, dx = \int |\cos{x}-\sin{x}| \, dx = \int |\sin{x}-\cos{x}| \, dx\)

which will be \(\displaystyle \sin{x}+\cos{x} + C\) when \(\displaystyle \cos{x} \ge \sin{x}\), and \(\displaystyle -(\cos{x}+\sin{x}) + C\) when \(\displaystyle \sin{x} \ge \cos{x}\)

 

[stapel]

∫[(1-sin2x)^1/2]dx = ∫ [(sin^2x + cos^2x - 2sinxcosx)^1/2] dx= ∫[(cosx - sinx)^2^1/2] dx = sinx + cosx + C

I'm supposed to get sinx - cosx + C

Typo in the solution manual, maybe?

sin^2x + cos^2x - 2sinxcosx = (sinx - cosx)^2 or (cosx - sinx)^2?

Not sure what you're doing here...? Note: Since you're squaring, the order in the subtraction doesn't matter. In reversing the subtraction, a "minus" sign is kicked out front, but it'll be squared to be +1, so it doesn't matter.