What is the integral of arccos(z) ?

The integral of arccos(z) is zarccos(z)-sqrt(1-z^2)+C

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integral of arccos(z)

\int arccos (z) d z

z arccos (z) - 1 - z^{2} + C

Solution steps

\int arccos (z) d z

Apply Integration By Parts

: z arccos (z) - \int - \frac{z}{1 - z ^{2}} d z

= z arccos (z) - \int - \frac{z}{1 - z ^{2}} d z

\int - \frac{z}{1 - z ^{2}} d z = 1 - z^{2}

= z arccos (z) - 1 - z^{2}

Add a constant to the solution

= z arccos (z) - 1 - z^{2} + C

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What is the integral of arccos(z) ?
The integral of arccos(z) is zarccos(z)-sqrt(1-z^2)+C