Evaluate the integral: $\int x \sec^2(8x) \, dx$

Neetesh Kumar | November 11, 2024

Calculus Homework Help

This is the solution to DHW Calculus
Assignment: 3 Question Number 10
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Step-by-step solution:

Step 1: Set up the integration by parts formula

We will use integration by parts, which is defined as:

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

We need to decide which parts of the integrand will be assigned to $u$ and $dv$.

• Let $u = x$ so that $du = dx$.
• Let $dv = \sec^2(8x) \, dx$. To integrate $dv$, we need to find $v$.

Step 2: Find the integral of $dv$

We know that the integral of $\sec^2(x)$ is $\tan(x)$, so we need to account for the factor of $8$ inside the function:

$v = \int \sec^2(8x) \, dx = \frac{1}{8} \tan(8x)$

This is because we have to divide by the inner derivative, which is $8$.

Step 3: Apply the integration by parts formula

Now, we apply the integration by parts formula:

$\int x \sec^2(8x) \, dx = x \cdot \frac{1}{8} \tan(8x) - \int \frac{1}{8} \tan(8x) \, dx$

Simplifying:

$= \frac{x}{8} \tan(8x) - \frac{1}{8} \int \tan(8x) \, dx$

Step 4: Solve the remaining integral

We now need to integrate $\tan(8x)$. The integral of $\tan(x)$ is $-\ln|\cos(x)|$, so we have:

$\int \tan(8x) \, dx = -\frac{1}{8} \ln|\cos(8x)|$

Step 5: Substitute the result back

Substitute the result into the expression:

$\frac{x}{8} \tan(8x) - \frac{1}{8} \left( -\frac{1}{8} \ln|\cos(8x)| \right)$

Simplify:

$= \frac{x}{8} \tan(8x) + \frac{1}{64} \ln|\cos(8x)| + C$

Where $C$ is the constant of integration.

Final Answer:

The value of the integral is:

$\boxed{\frac{x}{8} \tan(8x) + \frac{1}{64} \ln|\cos(8x)| + C}$

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