Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (anonymous):

Integrate arccos(x/2)

OpenStudy (anonymous):

\[\int{cos^{-1}\left(\frac{x}{2}\right)}dx=\int{cos^{-1}(u)}dx\] \[u=\frac{x}{2}\rightarrow du=\frac{1}{2}\rightarrow2du=dx\] \[\int{cos^{-1}(u)}dx=\int{2cos^{-1}(u)du}=2\int{cos^{-1}(u)}du=2\left[ucos^{-1}(u)-\sqrt{1-u^2}+C\right]\] \[=2\left[\frac{x}{2}cos^{-1}\left(\frac{x}{2}\right)-\sqrt{1-\left(\frac{x}{2}\right)^2}+C\right]=xcos^{-1}\frac{x}{2}-\sqrt{1-\frac{x^2}{4}}+2C\]

OpenStudy (anonymous):

err actually take the co-efficient off the \(C\) haha

OpenStudy (anonymous):

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

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!