SOLUTION 14: $\ \$ Integrate $\displaystyle{ \int { x \cdot \sqrt{10x-x^2} } \ dx }$. First complete the square. $$= \displaystyle { \int x \cdot \sqrt{10x-x^{2}} \ dx = \int x \cdot \sqrt{-(x^{2}-10x)} \ dx }$$ $$= \displaystyle { \int x \cdot \sqrt{-(x^{2}-10x+25)+25} \ dx }$$ $$= \displaystyle { \int x \cdot \sqrt{-(x-5)^{2}+25} \ dx }$$ Now use the trig substitution $$x-5 = 5 \sin \theta$$ so that $$dx = 5 \cos \theta \ d\theta$$ Substitute into the original problem, replacing all forms of x, getting $$\displaystyle { \int x \cdot \sqrt{-(x-5)^{2}+25} \ dx } = \displaystyle { \int (5 + 5 \sin \theta) \cdot \sqrt {-25 \sin^{2} \theta + 25} \cdot 5 \cos \theta \ d \theta }$$ $$= \displaystyle { \int (5 + 5 \sin \theta) \cdot \sqrt {25 (1-\sin^{2} \theta) } \cdot 5 \cos \theta \ d \theta }$$ $$= \displaystyle { \int (5 + 5 \sin \theta) \cdot \sqrt{25 \cos^{2} \theta} \cdot 5 \cos \theta \ d \theta }$$ $$= \displaystyle { \int 5(1+ \sin \theta) \cdot 5 \cos \theta \cdot 5 \cos \theta \ d \theta }$$ $$= \displaystyle { 125 \int (1 + \sin \theta) \cos^{2} \theta \ d \theta }$$ $$= \displaystyle { 125 \int (\cos^{2} \theta + \sin \theta \cos^{2} \theta) \ d \theta }$$ $$= \displaystyle { 125 \int \cos^{2} \theta \ d \theta + 125 \int \sin \theta \cos^{2} \theta \ d \theta }$$ $\Bigg($ Let $A = \displaystyle { \int \cos^{2} \theta \ d \theta } \$ and recall that $\ \cos 2 \theta = 2 \cos^{2} \theta - 1 \$ so that $\cos^{2} \theta = (1/2) \cos 2 \theta \$. Then $$\displaystyle { A = \int \frac{\cos 2 \theta + 1}{2} \ d \theta }$$ $$= \displaystyle { \int \Big( \frac{1}{2} \cos 2 \theta + \frac{1}{2} \Big) \ d \theta }$$ $$= \displaystyle { \frac{1}{2} \frac{\sin 2 \theta}{2} + \frac{\theta}{2} + C}$$ $$= \displaystyle { \frac{\sin 2 \theta}{4} + \frac{\theta}{2} + C}$$ Let $B = \int \sin \theta \cos^{2} \theta \ d \theta$. $\$ Use the ordinary u-ubstitution $$u = \cos \theta$$ so that $$du = - \sin \theta \ d \theta \ \ \longrightarrow \ \ - du = \sin \theta \ d \theta$$ Substitute into the original problem, replacing all forms of $u$, getting $$\displaystyle { \int \sin \theta \cos^{2} \theta \ d \theta = \int -u^{2} \ d \theta }$$ $$= \displaystyle { - \frac{u^{3}}{3} + C}$$ $$= \displaystyle { - \frac{\cos^{3} \theta}{3} + C } \ \Bigg)$$ $$= 125 A + 125 B$$ $$= \displaystyle { 125 \Big( \frac{\sin 2 \theta}{4} + \frac{\theta}{2} \Big) - 125 \Big(- \frac{\cos^{3} \theta}{3} \Big)+ C }$$ (Recall that $\sin 2 \theta = 2 \sin \theta \cos \theta$. ) $$= \displaystyle { \frac{125}{4} (2 \sin \theta \cos \theta) + \frac{125}{2} \theta + \frac{125}{3} \cos^{3} \theta + C }$$ $$= \displaystyle { \frac{125}{2} \sin \theta \cos \theta + \frac{125}{2} \theta + \frac{125}{3} \cos^{3} \theta + C }$$ $\Big($ We need to write our final answer in terms of $x$.

Since $\ x-5 = 5 \sin \theta \$ it follows that $\ \theta = \arcsin \displaystyle{ x-5 \over 5 } \$ and $$\sin \theta = \displaystyle{ x-5 \over 5 } = \displaystyle{ opposite \over hypotenuse }$$ and from the Pythagorean Theorem that $$\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(adjacent)^2 + (x-5)^2 = (5)^2 \ \ \longrightarrow \ \ \ adjacent = \sqrt{25-(x-5)^2} = \sqrt{10x-x^2} \ \ \longrightarrow$$ $$\cos \theta = \displaystyle{ adjacent \over hypotenuse }= \displaystyle{ \sqrt{10x-x^2} \over 5 }. \Big)$$ $$= \displaystyle { \frac{125}{2} \cdot \frac{x-5}{5} \cdot \frac{\sqrt{10x-x^2}}{5} + \frac{125}{2} \cdot \arcsin \frac{x-5}{10} - \frac{125}{3}\Big(\frac{\sqrt{10x-x^2}}{5} \Big)^{3} + C }$$ $$= \displaystyle { \frac{5}{2} (x-5) \sqrt{10x-x^2} + \frac{125}{2} \arcsin \frac{x-5}{10} - \frac{1}{3} (10x-x^2)^{3/2} + C }$$