Question

Evaluate the integral: $ \int_0^\pi \frac{\tan x}{\cos x} \, dx = \int_0^\pi \frac{\sin x}{\cos^2 x} \, dx $

• A. \( \infty \)
• B. \( 0 \)
• C. \( 1 \)
• D. Undefined

Hint:

When dealing with integrals involving trigonometric identities, substitution can often help simplify the problem and make it more straightforward to solve. Keep an eye on standard identities, such as \( \tan x = \frac{\sin x}{\cos x} \), to facilitate the calculation.

Answer 1

The Correct Option is A

We are given the integral: \[ \int_0^\pi \frac{\tan x}{\cos x} \, dx = \int_0^\pi \frac{\sin x}{\cos^2 x} \, dx \] First, observe that: \[ \tan x = \frac{\sin x}{\cos x} \] So, the original integral becomes: \[ \int_0^\pi \frac{\sin x}{\cos^2 x} \, dx \] Now, we need to evaluate this integral. To simplify this, we perform the substitution: \[ u = \cos x, \quad \frac{du}{dx} = -\sin x \]
Thus, the integral becomes: \[ \int_{u(0)}^{u(\pi)} \frac{-1}{u^2} \, du \] At the limits \( x = 0 \) and \( x = \pi \), we know that: - \( \cos 0 = 1 \) - \( \cos \pi = -1 \) Therefore, the integral becomes: \[ \int_1^{-1} \frac{-1}{u^2} \, du \] This simplifies to: \[ - \left[ \frac{1}{u} \right]_1^{-1} = - \left( \frac{1}{-1} - \frac{1}{1} \right) = -(-1 - 1) = 2 \] Hence, the answer is \( 2 \).