Dx/(sinx-cosx)$$ \frac{1}{\left(\sin x-\cos x\right)} $$

To solve the integral $\int \frac{dx}{\sin x - \cos x}$, we'll use a substitution method.

### Step 1: Simplify the expression
The expression inside the integral is $\frac{1}{\sin x - \cos x}$. To simplify it, we can multiply the numerator and denominator by $\frac{1}{\sqrt{2}}$ to get:

\[
\frac{1}{\sin x - \cos x} = \frac{1}{\sqrt{2}} \cdot \frac{1}{\frac{1}{\sqrt{2}}(\sin x - \cos x)}
\]

This simplifies to:

\[
\frac{1}{\sin x - \cos x} = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sin\left(x - \frac{\pi}{4}\right)}
\]

Thus, the integral becomes:

\[
\int \frac{dx}{\sin x - \cos x} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sin\left(x - \frac{\pi}{4}\right)}
\]

### Step 2: Use a substitution
Let $u = x - \frac{\pi}{4}$, so $du = dx$. The integral now simplifies to:

\[
\frac{1}{\sqrt{2}} \int \frac{du}{\sin u}
\]

### Step

Learn how to solve the integral of 1 divided by sine x minus cosine x using trigonometric substitution and integration techniques. This problem involves advanced mathematical concepts in integration and trigonometry.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation