Question

The value of \(\int_{0}^{\pi/3}\frac{\tan t}{\cos t} dt\) is equal to

• A. \(\frac{1}{2}\)
• B. \(-\frac{1}{2}\)
• C. 2
• D. -2
• E. 1

Hint:

\(\frac{\tan x}{\cos x} = \tan x \sec x\) and \(\frac{d}{dx}(\sec x) = \sec x \tan x\).

Answer 1

Answer: (E)

Step 1: Concept:
• Simplify the integrand: \[ \frac{\tan t}{\cos t} = \frac{\sin t}{\cos t} \cdot \frac{1}{\cos t} = \frac{\sin t}{\cos^2 t} \]

Step 2: Detailed Explanation:
• Rewrite the integral: \[ \int_{0}^{\pi/3} \frac{\tan t}{\cos t} \, dt = \int_{0}^{\pi/3} \frac{\sin t}{\cos^2 t} \, dt \]
• Let: \[ u = \cos t \Rightarrow du = -\sin t \, dt \]
• Change limits: \[ t = 0 \Rightarrow u = 1,\quad t = \frac{\pi}{3} \Rightarrow u = \frac{1}{2} \]
• Substitute: \[ \int_{0}^{\pi/3} \frac{\sin t}{\cos^2 t} \, dt = \int_{1}^{1/2} \frac{-du}{u^2} \]
• Reverse limits: \[ = \int_{1/2}^{1} u^{-2} \, du \]
• Integrate: \[ = \left[-u^{-1}\right]_{1/2}^{1} \]
• Evaluate: \[ = (-1) - (-2) = 1 \]

Step 3: Final Answer:
• The value is \(1\).