evaluate ∫ x^10 / (16 - x^2)^(13/2) dx.

# Explanation: ## Step1: Use trig - substitution Let $x = 4\sin\theta$, then $dx=4\cos\theta d\theta$. And $16 - x^{2}=16 - 16\sin^{2}\theta = 16\cos^{2}\theta$. ## Step2: Substitute into the integral The integral $\int\frac{x^{10}}{(16 - x^{2})^{\frac{13}{2}}}dx=\int\frac{(4\sin\theta)^{10}}{(16\cos^{2}\theta)^{\frac{13}{2}}}\cdot4\cos\theta d\theta$. Simplify the integrand: $(4\sin\theta)^{10}=4^{10}\sin^{10}\theta$, $(16\cos^{2}\theta)^{\frac{13}{2}}=16^{\frac{13}{2}}\cos^{13}\theta = 4^{13}\cos^{13}\theta$. The integral becomes $\int\frac{4^{10}\sin^{10}\theta}{4^{13}\cos^{13}\theta}\cdot4\cos\theta d\theta=\frac{1}{4^{2}}\int\frac{\sin^{10}\theta}{\cos^{12}\theta}d\theta=\frac{1}{16}\int\tan^{10}\theta\sec^{12}\theta d\theta$. ## Step3: Use the identity $\sec^{2}\theta = 1+\tan^{2}\theta$ Let $u = \tan\theta$, then $du=\sec^{2}\theta d\theta$. And $\sec^{12}\theta=(\sec^{2}\theta)^{6}=(1 + \tan^{2}\theta)^{6}$. The integral $\frac{1}{16}\int\tan^{10}\theta\sec^{12}\theta d\theta=\frac{1}{16}\int u^{10}(1 + u^{2})^{6}du$. Expand $(1 + u^{2})^{6}$ using the binomial theorem $(a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}$, where $a = 1$, $b=u^{2}$, $n = 6$. $(1 + u^{2})^{6}=\sum_{k = 0}^{6}\binom{6}{k}(u^{2})^{k}=1 + 6u^{2}+15u^{4}+20u^{6}+15u^{8}+6u^{10}+u^{12}$. Then $\frac{1}{16}\int u^{10}(1 + u^{2})^{6}du=\frac{1}{16}\int u^{10}(1 + 6u^{2}+15u^{4}+20u^{6}+15u^{8}+6u^{10}+u^{12})du$. $=\frac{1}{16}\int(u^{10}+6u^{12}+15u^{14}+20u^{16}+15u^{18}+6u^{20}+u^{22})du$. ## Step4: Integrate term - by - term $\int(u^{10}+6u^{12}+15u^{14}+20u^{16}+15u^{18}+6u^{20}+u^{22})du=\frac{u^{11}}{11}+\frac{6u^{13}}{13}+\frac{15u^{15}}{15}+\frac{20u^{17}}{17}+\frac{15u^{19}}{19}+\frac{6u^{21}}{21}+\frac{u^{23}}{23}+C$. $=\frac{u^{11}}{11}+\frac{6u^{13}}{13}+u^{15}+\frac{20u^{17}}{17}+\frac{15u^{19}}{19}+\frac{2u^{21}}{7}+\frac{u^{23}}{23}+C$. ## Step5: Back - substitute Since $u = \tan\theta$ and $x = 4\sin\theta$, $\sin\theta=\frac{x}{4}$, $\cos\theta=\frac{\sqrt{16 - x^{2}}}{4}$, $\tan\theta=\frac{x}{\sqrt{16 - x^{2}}}$. The final result is $\frac{1}{16}\left[\frac{\left(\frac{x}{\sqrt{16 - x^{2}}}\right)^{11}}{11}+\frac{6\left(\frac{x}{\sqrt{16 - x^{2}}}\right)^{13}}{13}+\left(\frac{x}{\sqrt{16 - x^{2}}}\right)^{15}+\frac{20\left(\frac{x}{\sqrt{16 - x^{2}}}\right)^{17}}{17}+\frac{15\left(\frac{x}{\sqrt{16 - x^{2}}}\right)^{19}}{19}+\frac{2\left(\frac{x}{\sqrt{16 - x^{2}}}\right)^{21}}{7}+\frac{\left(\frac{x}{\sqrt{16 - x^{2}}}\right)^{23}}{23}\right]+C$. # Answer: $\frac{1}{16}\left[\frac{x^{11}}{11(16 - x^{2})^{\frac{11}{2}}}+\frac{6x^{13}}{13(16 - x^{2})^{\frac{13}{2}}}+\frac{x^{15}}{(16 - x^{2})^{\frac{15}{2}}}+\frac{20x^{17}}{17(16 - x^{2})^{\frac{17}{2}}}+\frac{15x^{19}}{19(16 - x^{2})^{\frac{19}{2}}}+\frac{2x^{21}}{7(16 - x^{2})^{\frac{21}{2}}}+\frac{x^{23}}{23(16 - x^{2})^{\frac{23}{2}}}\right]+C$