Integrate: dx/sqrt[1 - x^2], for all x from -1 to 1. Please help and thanks.
My answer is the same as Heureka's, I just like to start it a little different.
\(\int_{-1}^1 \frac{1}{\sqrt{1 - x^2}}\;dx\)
I like to start with this little triangle it wasn't meant to be that big. 
\(cos\theta=\sqrt{1-x^2}\\~\\ sin\theta=x\\ x=\sin\theta\\ \frac{dx}{d\theta}=cos\theta\\ dx=cos\theta\;d\theta\\~\\ When\;\;x=1,\;\;\;\;sin\theta =1\;\;\;\;\;\;\;\;\rightarrow\;\;\;\theta=\frac{\pi}{2}\\ When\;\;x=-1,\;\;\;\;sin\theta =-1\;\;\;\;\rightarrow\;\;\;\theta=\frac{-\pi}{2}\\ SO\\ \)
\(\displaystyle \int_{-1}^1 \frac{1}{\sqrt{1 - x^2}}\;dx\\~\\ =\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\;\;\; \frac{1}{cos\theta}\;cos\theta\;d\theta\\~\\ =\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\;\;\; 1\;d\theta\\~\\ =\left[\theta\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\\~\\ =\frac{\pi}{2}-\;-\frac{\pi}{2}\\~\\ =\pi\)
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+118728 
