Problem 13 - Essential Calculus

Question

Use integration tables to find the integral. $$ \int e^{x} \arccos e^{x} d x $$

Step-by-Step Solution

Answer

The integral of \(e^x \arccos(e^x)\) with respect to 'x' is \(e^x \arccos(e^x) + e^x \sqrt{1-e^{2x}} - \arcsin(e^x) + C\)

Step 1: Set up the Integrand with 'u' and 'dv'

We choose 'u' and 'dv' such that 'u' becomes simpler after differentiation and 'dv' is easy to integrate. In this case, let \(u = \arccos e^x\) and \(dv = e^x dx\). Thus, compute \(du = \frac{-e^x dx}{\sqrt{1-e^{2x}}}\) and \(v = e^x\).

Step 2: Apply Integration by Parts Formula

Substituting 'u', 'v', 'du', and 'dv' into integration by parts formula \(\int udv = uv - \int vdu\), we get: \(\int e^x \arccos(e^x) dx = e^x \arccos(e^x) - \int e^x \frac{-e^x dx}{\sqrt{1-e^{2x}}}\). This simplifies to: \(e^x \arccos(e^x) + \int \frac{e^{2x}}{\sqrt{1-e^{2x}}} dx\)

Step 3: Simplify the Integral

The integral in the latest expression doesn't have a standard integral form yet. It can be replaced by a standard integral form by substituting \(u = e^x\) and \(du = e^x dx\), and it simplifies to \(\int \frac{u^2 du} {\sqrt{1-u^{2}}}\). This integral can now be found in the table of standard integrals.

Step 4: Evaluate the Simplified Integral Using the Table

Referring to the integral table, we find that \(\int \frac{u^2 du} {\sqrt{1-u^{2}}} = u \sqrt{1-u^{2}} - \arcsin(u) + C\). Substituting back \(u = e^x\), we get \(e^x \sqrt{1-e^{2x}} - \arcsin(e^x) + C\).

Step 5: Combine All the Parts to Get the Final Answer

Concatenating the previous steps, the solution of the original integral is \(e^x \arccos(e^x) + e^x \sqrt{1-e^{2x}} - \arcsin(e^x) + C\)