Calc1.66

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\int {\frac  {1}{{\sqrt  {4x-x^{2}}}}}\,dx\,

If we could get rid of the square root, we could probably do this integral. Let's take advantage of the identity that tells us

\cos ^{2}x=1-\sin ^{2}x\,

First note that

\int {\frac  {1}{{\sqrt  {4x-x^{2}}}}}\,dx=\int {\frac  {1}{{\sqrt  {4-(x-2)^{2}}}}}\,dx\,

Keeping that in mind, let us use the substitution

x-2=2\sin \theta \,

dx=2\cos \theta \,d\theta \,

This gives us

\int {\frac  {1}{{\sqrt  {4-(x-2)^{2}}}}}\,dx=\int {\frac  {2\cos \theta }{{\sqrt  {4-4\sin ^{2}\theta }}}}d\theta =\int {\frac  {\cos \theta }{\cos \theta }}d\theta =\theta +C=\arcsin {\frac  {x-2}{2}}+C\,


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