Chapter 6.3, Problem 26E

To determine

To find: The integral analytically.

Answer to Problem 26E

The answer is: $\frac{3}{4}-\frac{3{\pi }^2}{16}\approx -1.101$

Explanation of Solution

Given information:

The integration is: ${{\displaystyle \int }}_0^{\pi /2}{x}^3\cos 2xdx$

Calculation:

The given integration is: ${{\displaystyle \int }}_0^{\pi /2}{x}^3\cos 2xdx$

Use tabular integration with

  $\begin{array}{c}f(x)=x^3\\ g(x)=\cos 2x\end{array}$

Then, discover the anti derivatives of $\cos 2x$ and the derivatives of $x^3$ until you reach 0. Then, draw arrows connecting each first-column expression (except from the final expression of 0) to the expression in the second-column row below. Start with a positive sign and alternate between positive and negative signs as write the signs above the arrows.

  

Utilize the operation signals located above each arrow to combine the results of the functions that are connected by arrows.

  $\begin{array}{c}{{\displaystyle \int }}_0^{\pi /2}{x}^3\cos 2xdx={\left.\left[x^3\left(\frac{1}{2}\sin 2x\right)-3x^2\left(-\frac{1}{4}\cos 2x\right)+6x\left(-\frac{1}{8}\sin 2x\right)-6\left(\frac{1}{16}\cos 2x\right)\right]\right|}_0^{\pi /2}\\ ={\left.\left[\frac{1}{2}{x}^3\sin 2x+\frac{3}{4}{x}^2\cos 2x-\frac{3}{4}x\sin 2x-\frac{3}{8}\cos 2x\right]\right|}_0^{\pi /2}\\ ={\left.\left[\left(\frac{1}{2}{x}^3-\frac{3}{4}x\right)\sin 2x+\left(\frac{3}{4}{x}^2-\frac{3}{8}\right)\cos 2x\right]\right|}_0^{\pi /2}\\ ={\left.\left[\frac{1}{4}\left(2x^3-3x\right)\sin 2x+\frac{3}{8}\left(2x^2-1\right)\cos 2x\right]\right|}_0^{\pi /2}\end{array}$

To calculate the integral's value, replace:

  $\begin{array}{c}\left[\frac{1}{4}\left(2{\left(\frac{\pi }{2}\right)}^3-3\left(\frac{\pi }{2}\right)\right)\sin \left(2\cdot \frac{\pi }{2}\right)+\frac{3}{8}\left(2{\left(\frac{\pi }{2}\right)}^2-1\right)\cos \left(2\cdot \frac{\pi }{2}\right)\right]-\left[\frac{1}{4}\left(2{(0)}^3-3(0)\right)\sin (2\cdot 0)+\frac{3}{8}\left(2{(0)}^2-1\right)\cos (2\cdot 0)\right]\\ =\left[\frac{1}{4}\left(2\cdot \frac{{\pi }^3}{8}-\frac{3\pi }{2}\right)\sin \pi +\frac{3}{8}\left(2\cdot \frac{{\pi }^2}{4}-1\right)\cos \pi \right]-\left[\frac{1}{4}(0-0)\sin 0+\frac{3}{8}(0-1)\cos 0\right]\\ =\left[\frac{1}{4}\left(\frac{{\pi }^3}{4}-\frac{3\pi }{2}\right)(0)+\frac{3}{8}\left(\frac{{\pi }^2}{2}-1\right)(-1)\right]-\left[0+\frac{3}{8}(-1)(1)\right]\\ =\left[0-\frac{3{\pi }^2}{16}+\frac{3}{8}\right]+\frac{3}{8}\\ =\frac{3}{4}-\frac{3{\pi }^2}{16}\end{array}$

To use a calculator gives $\frac{3}{4}-\frac{3{\pi }^2}{16}\approx -1.101$

To use NINT to find ${{\displaystyle \int }}_0^{\pi /2}{x}^3\cos 2xdx$ also gives about $-1.101$ so the answer is correct.

Therefore the required integral analytically of the given integration ${{\displaystyle \int }}_0^{\pi /2}{x}^3\cos 2xdx$ is

  $\frac{3}{4}-\frac{3{\pi }^2}{16}\approx -1.101$