Question

$\int_{}^{}{\mathbf{\tan}\mathbf{(}\mathbf{3x}\mathbf{-}\mathbf{5)}\mathbf{\sec}\mathbf{(}\mathbf{3x}\mathbf{-}\mathbf{5)dx =}}$

• A. $\sec(3x - 5) + c$
• B. $\frac{1}{3}\sec(3x - 5) + c$
• C. $\tan(3x - 5) + c$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{3}\sec(3x - 5) + c$

Answer 2

$\because\int_{}^{}{\sec x\tan xdx = \sec x + c}$

$\therefore\int_{}^{}{\sec(3x - 5)\tan(3x - 5)dx}$ $= \frac{\sec(3x - 5)}{3} + C$