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# Explanation: ## Step1: Rewrite $\sec^{6}(x)$ Since $\sec^{2}(x)=1 + \tan^{2}(x)$, then $\sec^{6}(x)=\sec^{2}(x)\cdot\sec^{2}(x)\cdot\sec^{2}(x)=(1 + \tan^{2}(x))^{3}$. So the integral becomes $\int16\tan^{4}(x)(1 + \tan^{2}(x))^{3}dx$. Let $u = \tan(x)$, then $du=\sec^{2}(x)dx$. ## Step2: Expand $(1 + u^{2})^{3}$ Using the binomial - expansion formula $(a + b)^{n}=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}$, for $(1 + u^{2})^{3}=1 + 3u^{2}+3u^{4}+u^{6}$. ## Step3: Multiply by $u^{4}$ $16u^{4}(1 + 3u^{2}+3u^{4}+u^{6})=16u^{4}+48u^{6}+48u^{8}+16u^{10}$. ## Step4: Integrate term - by - term $\int(16u^{4}+48u^{6}+48u^{8}+16u^{10})du=16\int u^{4}du+48\int u^{6}du+48\int u^{8}du+16\int u^{10}du$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have: $16\times\frac{u^{5}}{5}+48\times\frac{u^{7}}{7}+48\times\frac{u^{9}}{9}+16\times\frac{u^{11}}{11}+C$. $=\frac{16}{5}u^{5}+\frac{48}{7}u^{7}+\frac{16}{3}u^{9}+\frac{16}{11}u^{11}+C$. ## Step5: Substitute back $u=\tan(x)$ $\frac{16}{5}\tan^{5}(x)+\frac{48}{7}\tan^{7}(x)+\frac{16}{3}\tan^{9}(x)+\frac{16}{11}\tan^{11}(x)+C$. # Answer: $\frac{16}{5}\tan^{5}(x)+\frac{48}{7}\tan^{7}(x)+\frac{16}{3}\tan^{9}(x)+\frac{16}{11}\tan^{11}(x)+C$