$$ \int _{}^{}x(\cos (2x))^{-1} d x $$

Question

UpStudy AI · Accepted Answer

To evaluate the integral:

$$
\int \frac{x}{\cos(2x)} \, dx = \int x \sec(2x) \, dx
$$

we can use **integration by parts**. Recall the integration by parts formula:

$$
\int u \, dv = uv - \int v \, du
$$

Let's choose:
- $$ u = x $$  ⇒ $$ du = dx $$
- $$ dv = \sec(2x) \, dx $$ ⇒ $$ v = \frac{1}{2} \ln |\sec(2x) + \tan(2x)| + C $$

Applying integration by parts:

$$
\int x \sec(2x) \, dx = x \cdot \frac{1}{2} \ln |\sec(2x) + \tan(2x)| - \int \frac{1}{2} \ln |\sec(2x) + \tan(2x)| \, dx + C
$$

Simplifying, we get:

$$
\int x \sec(2x) \, dx = \frac{x}{2} \ln |\sec(2x) + \tan(2x)| - \frac{1}{2} \int \ln |\sec(2x) + \tan(2x)| \, dx + C
$$

### Handling the Remaining Integral

The remaining integral:

$$
\int \ln |\sec(2x) + \tan(2x)| \, dx
$$

does not have a simple expression in terms of elementary functions. Therefore, the final expression for the original integral is:

$$
\int \frac{x}{\cos(2x)} \, dx = \frac{x}{2} \ln |\sec(2x) + \tan(2x)| - \frac{1}{2} \int \ln |\sec(2x) + \tan(2x)| \, dx + C
$$

where $$ C $$ is the constant of integration.

**Conclusion:** The integral cannot be expressed using elementary functions beyond the expression obtained through integration by parts. Thus, the final answer involves an unevaluated integral of a logarithmic function.

**Final Answer:**

After expanding by parts, an antiderivative is (½)x ln|sec 2x + tan 2x| minus half the integral of ln(sec 2x + tan 2x) dx. Thus,

∫x cos⁻¹(2x) dx = (x⁄2) ln|sec2x + tan2x| − (½) ∫ln(sec2x + tan2x) dx + C
The integral cannot be expressed using elementary functions beyond the expression obtained through integration by parts. Thus, $$ \int \frac{x}{\cos(2x)} \, dx = \frac{x}{2} \ln |\sec(2x) + \tan(2x)| - \frac{1}{2} \int \ln |\sec(2x) + \tan(2x)| \, dx + C $$ where $$ C $$ is the constant of integration.

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Handling the Remaining Integral

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