Question:medium

Evaluate: \[ \int_{0}^{\pi/4} \frac{1}{\sin x + \cos x} \, dx \]

Show Hint

For definite integrals involving trigonometric terms, simplify the denominator and use substitution for ease.
Updated On: Jan 13, 2026
Show Solution

Solution and Explanation

1. Simplify denominator: Use the identity \[\sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right).\] This yields \[\frac{1}{\sin x + \cos x} = \frac{1}{\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)}.\] 2. Rewrite integral: \[\int_{0}^{\pi/4} \frac{1}{\sin x + \cos x} \, dx = \frac{1}{\sqrt{2}} \int_{0}^{\pi/4} \csc\left(x + \frac{\pi}{4}\right) \, dx.\] 3. Substitute: Let \( u = x + \frac{\pi}{4} \). Then \( du = dx \). Adjust the limits: \( u = \frac{\pi}{4} \to \frac{\pi}{2} \) and \( u = 0 \to \frac{\pi}{4} \). The integral becomes \[\frac{1}{\sqrt{2}} \int_{\pi/4}^{\pi/2} \csc u \, du.\] 4. Integrate \( \csc u \): \[\int \csc u \, du = \ln|\csc u - \cot u|.\] 5. Evaluate at limits: \[\frac{1}{\sqrt{2}} \left[\ln|\csc(\pi/2) - \cot(\pi/2)| - \ln|\csc(\pi/4) - \cot(\pi/4)|\right].\] Substitute values: \[\csc(\pi/2) = 1, \, \cot(\pi/2) = 0, \, \csc(\pi/4) = \sqrt{2}, \, \cot(\pi/4) = 1.\] Simplify: \[\ln|\csc(\pi/2) - \cot(\pi/2)| = \ln(1), \quad \ln|\csc(\pi/4) - \cot(\pi/4)| = \ln(\sqrt{2} - 1).\] The final result is \[\frac{1}{\sqrt{2}} \left[0 - \ln(\sqrt{2} - 1)\right] = -\frac{\ln(\sqrt{2} - 1)}{\sqrt{2}}.\] Final Answer: \[\boxed{-\frac{\ln(\sqrt{2} - 1)}{\sqrt{2}}.}.\]
Was this answer helpful?
0