Question

The integral $\int {\sec }^{\frac{2}{3}}xcose{c}^{\frac{4}{3}}xdx$ is equal to

• A. $-3{\cot }^{-\frac{1}{3}}x+C$
• B. $-3{\tan }^{-\frac{1}{3}}x+C$
• C. $-\frac{3}{4}{\tan }^{-\frac{4}{3}}x+C$
• D. $3{\tan }^{-\frac{1}{3}}x+C$

Answer 1

The correct option is B $-3{\tan }^{-\frac{1}{3}}x+C$

$I=\int {\sec }^{\frac{2}{3}}xcose{c}^{\frac{4}{3}}xdx=\int \frac{1}{{\cos }^{\frac{2}{3}}x{\sin }^{\frac{4}{3}}x}dx=\int \frac{dx}{(\frac{\sin x}{\cos x}{)}^{\frac{4}{3}}\cdot {\cos }^{2}x}=\int \frac{{\sec }^{2}x}{(\tan x{)}^{\frac{4}{3}}}dx\text{Let}\tan x=t\Rightarrow {\sec }^{2}xdx=dt\therefore I=\int \frac{dt}{t^{\frac{4}{3}}}=-\frac{3}{t^{\frac{1}{3}}}+C=-3{\tan }^{-\frac{1}{3}}x+C$