Integral

**Reminisce** · February 16th 2010, 05:34 PM

Attachment...

Just wondering if this is correct..

**Krizalid** · February 16th 2010, 05:45 PM

no, it's actually $\int_{0}^{\pi }{\sqrt{1-\cos ^{2}2x}\,dx}=\int_{0}^{\pi }{\left| \sin 2x \right|\,dx}.$

**pickslides** · February 16th 2010, 05:45 PM

It's not correct, use $1-\cos^2 \theta = \sin^2\theta$

**Reminisce** · February 16th 2010, 06:07 PM

that was actually the first thing I tried however I got 0
and the answer is suppose to be 2

is there some kind of rule when using absolute values??

**Krizalid** · February 16th 2010, 08:36 PM

$\int_{0}^{\pi }{\left| \sin 2x \right|\,dx}=\frac{1}{2}\int_{0}^{2\pi }{\left| \sin x \right|\,dx}=\frac{1}{2}\left( \int_{0}^{\pi }{\sin x\,dx}-\int_{\pi }^{2\pi }{\sin x\,dx} \right)=2.$

**Reminisce** · February 16th 2010, 08:50 PM

o.O what?!? where did the 1/2 and 2pi come from and how did you remove the 2 from sin(2x)...???

**Krizalid** · February 16th 2010, 09:45 PM

substitution $t=2x.$

**Reminisce** · February 16th 2010, 10:09 PM

oh okay yeah it took me awhile cause you kept sinx as sinx

but im still getting 0

-cos(pi) = 1
-cos(0) = -1

--cos(2pi) = +1
--cos(pi) = -1

1-1+1-1 = 0

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