Question

If $\int f\left(x\right)dx=\frac{3}{55}\sqrt[3]{3{\tan }^{5}x}(5{\tan }^{2}x+11)+C$ then $f\left(x\right)$ is equal to

• A. $\sqrt[3]{3{\sin }^{2}x{\cos }^{-14}x}$
• B. $\sqrt[3]{3{\tan }^{2}x(1+{\tan }^{2}x{)}^6}$
• C. $\sqrt[3]{3{\cos }^{2}x{\sin }^{-14}x}$
• D. $\frac{7}{3}\sqrt[3]{3{\sin }^{2}x{\cos }^{-14}x}$

Answer 1

Answer: (A), (B)

$\sqrt[3]{3{\sin }^{2}x{\cos }^{-14}x}$
$\sqrt[3]{3{\tan }^{2}x(1+{\tan }^{2}x{)}^6}$

$\int f\left(x\right)dx=\frac{3}{55}\sqrt[3]{3{\left(\tan x\right)}^5}\left(5{\tan }^{2}x+11\right)$
$f\left(x\right)=\frac{d}{dx}\left(\frac{3}{55}\sqrt[3]{3{\left(\tan x\right)}^5}\left(5{\tan }^{2}x+11\right)\right)$
$=\frac{3}{55}\frac{d}{dx}(5{\tan }^{\frac{11}{3}}x+11{\tan }^{\frac{5}{3}}x){\sec }^{2}x$
$=\frac{3}{55}(\frac{55}{3}{\tan }^{\frac{8}{3}}x+\frac{55}{3}{\tan }^{\frac{2}{3}}x){\sec }^{2}x$
$=({\tan }^{\frac{8}{3}}x+{\tan }^{\frac{2}{3}}x){\sec }^{2}x={\tan }^{\frac{2}{3}}x\left({(1+{\tan }^{2}x)}^2\right)=\sqrt[3]{3{\sin }^{2}x{\cos }^{-14}x}$