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Following from the previous question

Evaluating $\int \arccos\bigl(\frac{\cos(x)}{r}\bigr) \, \mathrm{d}x$

I now need the extra $\sin^2x$ as in the title. Of course one power of $\sin(x)$ is easy, but it's not clear that two can be done using parts, or the methods of

Indefinite integral $\int \arcsin \left(k\sin x\right) dx$

Any suggestions?

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Thanks to the quoted answer, it suffices to compute \begin{align}I&=\int \arccos \left(k\cos x\right)2\cos 2x \,dx=\\&=\sin 2x \arccos \left(k\cos x\right) -\int\frac{k\sin 2x\,\sin x\,dx}{\sqrt{1-k^2\cos^2 x}}=\\ &=\sin 2x \arccos \left(k\cos x\right)-2k\int\frac{\sin^2x\,d(\sin x)}{\sqrt{1-k^2\cos^2 x}}, \end{align} where the transition from the 1st to the 2nd line is obtained by integration by parts. The remaining integral $\int\frac{s^2ds}{\sqrt{1-k^2+k^2s^2}}$ is clearly expressible in terms of elementary functions, and the final result can be simplified to $$I=\sin 2x \arccos \left(k\cos x\right)+\frac{1-k^2}{k^2}\ln\left( \sqrt{1-k^2\cos^2x}+k\sin x\right)-\frac{\sin x\,\sqrt{1-k^2\cos^2x}}{k}.$$

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