To integrate $\int \frac{1}{\sin x + \cos x} \, dx$, we use substitution and trigonometric identities to simplify the expression.

Solution:

Step 1: Simplify the denominator using a substitution

We know: $\sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right),$ since $\sin x + \cos x$ can be rewritten using the angle addition formula.

Step 2: Substitution

Let: $t = x + \frac{\pi}{4}.$ This gives: $\sin x + \cos x = \sqrt{2} \sin t \quad \text{and} \quad dx = dt.$

The integral becomes: $\int \frac{1}{\sin x + \cos x} \, dx = \int \frac{1}{\sqrt{2} \sin t} \, dt = \frac{1}{\sqrt{2}} \int \csc t \, dt.$

Step 3: Integral of $\csc t$

The integral of $\csc t$ is: $\int \csc t \, dt = \ln|\csc t - \cot t| + C.$

Thus: $\frac{1}{\sqrt{2}} \int \csc t \, dt = \frac{1}{\sqrt{2}} \ln|\csc t - \cot t| + C.$

Step 4: Back-substitution

Return to the original variable $x$: $\csc t = \frac{1}{\sin t} = \frac{\sqrt{2}}{\sin(x + \pi/4)},$ and: $\cot t = \frac{\cos t}{\sin t} = \frac{\cos(x + \pi/4)}{\sin(x + \pi/4)}.$

Thus, the final answer is: $\int \frac{1}{\sin x + \cos x} \, dx = \frac{1}{\sqrt{2}} \ln\left|\frac{\sqrt{2}}{\sin(x + \pi/4)} - \frac{\cos(x + \pi/4)}{\sin(x + \pi/4)}\right| + C.$

This simplifies further into: $\int \frac{1}{\sin x + \cos x} \, dx = \frac{1}{\sqrt{2}} \ln\left|\frac{1 - \cos(x + \pi/4)}{\sin(x + \pi/4)}\right| + C.$

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Related Questions:

1. What is the integral of $\int \frac{1}{\sin x - \cos x} \, dx$?
2. How do trigonometric substitutions work in integrals?
3. Can this integral be solved by a different method?
4. How does the substitution $u = \tan(x/2)$ help solve integrals with trigonometric expressions?
5. What is the geometrical interpretation of $\sin x + \cos x$?

Tip:

When solving trigonometric integrals, identifying common identities like $\sin x + \cos x = \sqrt{2} \sin(x + \pi/4)$ simplifies the process greatly.