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Integrate arccos(x/2)
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\[\int{cos^{-1}\left(\frac{x}{2}\right)}dx=\int{cos^{-1}(u)}dx\] \[u=\frac{x}{2}\rightarrow du=\frac{1}{2}\rightarrow2du=dx\] \[\int{cos^{-1}(u)}dx=\int{2cos^{-1}(u)du}=2\int{cos^{-1}(u)}du=2\left[ucos^{-1}(u)-\sqrt{1-u^2}+C\right]\] \[=2\left[\frac{x}{2}cos^{-1}\left(\frac{x}{2}\right)-\sqrt{1-\left(\frac{x}{2}\right)^2}+C\right]=xcos^{-1}\frac{x}{2}-\sqrt{1-\frac{x^2}{4}}+2C\]
err actually take the co-efficient off the \(C\) haha
Actually correction haha: \[\int cos^{-1}\frac{x}{2}=xcos^{-1}\frac{x}{2}-2\sqrt{1-\frac{x^2}{4}}+C\] FINAL ANSWER haha
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