Question

Evaluate: \( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{1 - \sin x} \, dx \)

Hint:

When integrating trigonometric functions, look for identities that simplify the integrand, and use substitution when appropriate.

Answer 1

Step 1: Use the trigonometric identity.
We need to simplify the integrand. We can multiply the numerator and denominator by \( 1 + \sin x \) to rationalize the denominator: \[ \frac{1}{1 - \sin x} \times \frac{1 + \sin x}{1 + \sin x} = \frac{1 + \sin x}{(1 - \sin^2 x)} = \frac{1 + \sin x}{\cos^2 x} \]

Step 2: Simplify the integral.
Now the integral becomes: \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1 + \sin x}{\cos^2 x} \, dx \] This can be split into two parts: \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\cos^2 x} \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sin x}{\cos^2 x} \, dx \]

Step 3: Solve the first integral.
The first integral is: \[ \int \frac{1}{\cos^2 x} \, dx = \tan x \]

Step 4: Solve the second integral.
For the second integral, use substitution \( u = \cos x \), so \( du = -\sin x \, dx \): \[ \int \frac{\sin x}{\cos^2 x} \, dx = - \int \frac{du}{u^2} = \frac{1}{u} = \frac{1}{\cos x} \]

Step 5: Combine the results.
Now combine the results from both integrals: \[ \left[ \tan x + \frac{1}{\cos x} \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \] Evaluating the limits, we get: \[ \left( \tan \frac{\pi}{2} + \sec \frac{\pi}{2} \right) - \left( \tan \frac{\pi}{4} + \sec \frac{\pi}{4} \right) \] \[ = \left( \infty + 1 \right) - \left( 1 + \sqrt{2} \right) = \infty \]

Final Answer: The integral diverges, so the answer is: \[ \boxed{\infty} \]