Attachment... Just wondering if this is correct..
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no, it's actually $\displaystyle \int_{0}^{\pi }{\sqrt{1-\cos ^{2}2x}\,dx}=\int_{0}^{\pi }{\left| \sin 2x \right|\,dx}.$
It's not correct, use $\displaystyle 1-\cos^2 \theta = \sin^2\theta $
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??
$\displaystyle \int_{0}^{\pi }{\left| \sin 2x \right|\,dx}=\frac{1}{2}\int_{0}^{2\pi }{\left| \sin x \right|\,dx}=\frac{1}{2}\left( \int_{0}^{\pi }{\sin x\,dx}-\int_{\pi }^{2\pi }{\sin x\,dx} \right)=2.$
o.O what?!? where did the 1/2 and 2pi come from and how did you remove the 2 from sin(2x)...???
substitution $\displaystyle t=2x.$
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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