$\int \sqrt{25-x^2}x^2\,dx$ with cos-substitution

Question

I have tried this integral with cos substitution, but I don't understand why it's wrong. So could you please tell me which step is wrong? Here are my steps: $$ x=5\cos\theta\\ dx=-5\sin\theta\,d\theta\\ $$ therefore \begin{align} &(1) = \int \sqrt{25-25\cos^2θ} \, 25\cos^{2}\theta \, (-5\sin\theta) \, d\theta\\ &(2) = -5^4\int \sin^{2}\theta \, \cos^{2}\theta\, d\theta \\ &(\sin 2\theta=2\sin\theta\cos\theta \implies \sin^22\theta=4\sin^2\theta\cos^2\theta)\\ &(3) =\frac{-5^4}{4}\int\sin^{2}2\theta\,d\theta\\ &(\sin^2\theta=\frac{1}{2}(1-\cos2\theta) \implies \sin^22\theta=\frac{1}{2}(1-\cos4\theta)\\ &(4) =\frac{-5^4}{8}\int (1-\cos4\theta)\,d\theta\\ &(5) = \frac{-5^4}{8} \left(\theta-\frac{1}{4}\sin4\theta\right) + C \\ &(\sin4\theta=2\sin2\theta\cos2\theta=4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta))\\ &(6)=\frac{-5^4}{8}(\theta-\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)) + C\\ &(\cos\theta=\frac{x}{5} \implies \sin\theta=\frac{\sqrt{25-x^2}}{5})\\ &(7)= \frac{-5^4}{8}\left(\arccos\frac{x}{5}-\frac{\sqrt{25-x^2}}{5}\cdot\frac{x}{5}\cdot\frac{25-2x^2}{5^2}\right) + C\\ &(8)=\frac{-5^4}{8}\left(\arccos\frac{x}{5}-\frac{x\sqrt{25-x^2}}{25}\cdot\frac{25-2x^2}{25}\right) + C\\ \end{align}

Answer 1

It all looks fine, except for a little mistake here in line (7): $$\frac{-5^4}{8}\left(\arccos\frac{x}{5}-\frac{\sqrt{25-x^2}}{5}\cdot\frac{x}{5}\cdot\frac{\color{red}{25-2x^2}}{5^2}\right)+C,$$ where your sign is flipped. Instead that fraction must be $$\cos^2\theta-\sin^2\theta=\left(\frac{x}{5}\right)^2-\left(\frac{\sqrt{25-x^2}}{5}\right)^2=\frac{x^2}{25}-\frac{25-x^2}{25}=\frac{2x^2-25}{25}.$$ As far as I can tell, this is the only mistake.