Evaluate the integral: $\int \sin^{-1}(4x) \, dx$

Neetesh Kumar | November 11, 2024

Calculus Homework Help

This is the solution to DHW Calculus
Assignment: 3 Question Number 11
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Step-by-step solution:

We need to evaluate the integral of $\sin^{-1}(4x)$. To do this, we use integration by parts.

Step 1: Choose $u$ and $dv$

We start by setting:

• $u = \sin^{-1}(4x) \quad \text{so that} \quad du = \frac{4}{\sqrt{1-(4x)^2}} \, dx$
• $dv = dx \quad \text{so that} \quad v = x$

Step 2: Apply the integration by parts formula

The integration by parts formula is:

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

Substitute the values of $u$, $du$, $v$, and $dv$ into the formula:

$\int \sin^{-1}(4x) \, dx = x \sin^{-1}(4x) - \int \frac{4x}{\sqrt{1 - (4x)^2}} \, dx$

Step 3: Solve the remaining integral

To solve the remaining integral $\int \frac{4x}{\sqrt{1 - (4x)^2}} \, dx$, we use the substitution:

• Let $u = 1 - (4x)^2$, so that $du = -8x \, dx$

Substituting this into the integral:

$\int \frac{4x}{\sqrt{1 - (4x)^2}} \, dx = -\frac{1}{2} \int \frac{du}{\sqrt{u}}$

This is a standard integral, and its solution is:

$-\frac{1}{2} \cdot 2\sqrt{u} = -\sqrt{u} = -\sqrt{1 - (4x)^2}$

Step 4: Substitute the result back

Now substitute back into the expression for the original integral:

$\int \sin^{-1}(4x) \, dx = x \sin^{-1}(4x) + \sqrt{1 - (4x)^2} + C$

Where $C$ is the constant of integration.

Final Answer:

The value of the integral is:

$\boxed{x \sin^{-1}(4x) + \sqrt{1 - (4x)^2} + C}$

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