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The integral in question is this:

$$\frac{1}{2}\int_0^{2\pi} [\cos^{-1}(\sin x)]^2~dx$$

The Mathematica code is

Integrate[(Acos[Sin[x]])^2/2, {x, 0, 2 Pi}]

Attempting this with Mathematica appears to be fruitless. It won't even do a numerical integration. WolframAlpha, however, deftly provides

$$\frac{1}{2}\int_0^{2\pi} [\cos^{-1}(\sin x)]^2~dx=\frac{\pi^3}{3}\approx 10.335$$

However, WA also declares a step-by-step solution is unavailable.

Incidentally, I know that the WA solution is correct because I have verified it numerically.

Cye Waldman
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1 Answers1

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Well, because Mathematica is a programming language and has a defined syntax. The function is called ArcCos and not Acos:

Integrate[(ArcCos[Sin[x]])^2/2, {x, 0, 2 Pi}] 
(* Pi^3/3 *)
halirutan
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  • Thanks.I suspected that but it gave me the correct interpretation in writing out the integral with $\cos^{-1}$. – Cye Waldman Mar 31 '18 at 22:52