Question

In Exercises \(21-24,\) use tabular integration to find the antiderivative. $$\int x^{3} \cos 2 x d x$$

Short answer

The antiderivative of the function is \(\frac{1}{2} x^{3} \sin 2 x - 3 x^{2}\frac{1}{2}\sin 2x + 6x\frac{1}{2}\cos 2x - 6\frac{1}{2}\sin 2x + C\), where C is the constant of integration.

Step-by-step solution

Selection of 'u' and 'dv'

In the integral \(\int x^3 \cos 2x \) dx, let 'u' be \(x^3\) and 'dv' be \(\cos 2x \) dx. This is because a simpler integral can be obtained if \(x^3\) is differentiated compared to \(\cos 2x\).

Find 'du' and 'v'

Differentiate 'u' to find 'du', and integrate 'dv' to find 'v'. Therefore, 'du' is \(3x^2 dx\) and 'v' is \(\frac{1}{2} \sin 2x\).

Apply Tabular Integration

Write a table with 'u' on the top and 'dv' at the bottom. The idea is to repeatedly differentiate 'u' and integrate 'dv' until 'u' becomes zero. 'u' becomes zero after three differentiations, so the process stops there. Then, alternate between plus and minus signs for each row. Start with a plus and end with a minus. Now, multiply the terms diagonally.

Write the Answer

Add up the terms obtained from Step 3. The integral becomes \(\frac{1}{2} x^{3} \sin 2 x - 3 x^{2}\frac{1}{2}\sin 2x + 6x\frac{1}{2}\cos 2x - 6\frac{1}{2}\sin 2x\)