Integral of $\int x^2\sqrt{25-x^2}\,dx$ using 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: $$ x=5\cos\theta\\ dx=-5\sinθ\,dθ\\ $$ therefore \begin{align} &(1)=\int\sqrt{25-25\cos^2θ}\left(25\cos^2θ\right)(-5\sinθ)\,dθ\\ &(2)=-5^4\int\sin^2θ\cos^2θ\,dθ\\ &\qquad\left(\sin2θ=2\sinθ\cosθ\implies\sin^22θ=4\sin^2θ\cos^2θ\right)\\ &(3)=\frac{-5^4}4\int\sin^22θ\,dθ\\ &\qquad\left(\sin^2θ=\frac12(1-\cos2θ)\implies\sin^22θ=\frac12(1-\cos4θ)\right)\\ &(4)=\frac{-5^4}8\int(1-\cos4θ)\,dθ\\ &(5)=\frac{-5^4}8\left(θ-\frac14\sin4θ\right)+C\\ &\qquad\left(\sin4θ=2\sin2θ\cos2θ=4\sinθ\cosθ\left(\cos^2θ-\sin^2θ\right)\right)\\ &(6)=\frac{-5^4}8\left(θ-\sinθ\cosθ\left(\cos^2θ-\sin^2θ\right)\right)+C\\ &\qquad\left(\cosθ=\frac x5\implies\sinθ=\frac{\sqrt{25-x^2}}5\right)\\ &(7)=\frac{-5^4}8\left(\arccos\left(\frac x5\right)-\frac{\sqrt{25-x^2}}5*\frac x5*\frac{25-2x^2}{5^2}\right)+C\\ &(8)=\frac{-5^4}8\left(\arccos\left(\frac x5\right)-\frac{x\sqrt{25-x^2}}{25}*\frac{25-2x^2}{25}\right)+C\\ \end{align}

Answer 1

$-5^4\int \sin^2\theta\cos^2\theta \ d\theta \\ -\frac {5^4}{4}\int \sin^2 2\theta \ d\theta\\ -\frac {5^4}{8}\int 1 - \cos 4\theta \ d\theta\\ -\frac {5^4}{8}(\theta - \frac 14 \sin 4\theta) + C\\ -\frac {5^4}{8}(\arccos \frac {\theta}{5} - \frac 14 (2\sin 2\theta\cos 2\theta)) + C\\ -\frac {5^4}{8}(\arccos \frac {\theta}{5} - \sin \theta\cos\theta(2\cos^2\theta - 1)) + C\\ -\frac {5^4}{8}(\arccos \frac {\theta}{5} - \sqrt {1-\frac {x^2}{25}}(\frac {x}{5})(2\frac {x^2}{25} - 1)) + C\\ \frac {1}{8} (-5^4\arccos \frac {x}{5} + (\sqrt {25- x^2})(x)(2x^2 - 25)) + C\\ $