Question

Q. Solve each of the integrals in Exercises $27–70.$Some integrals require integration by parts, and some do not.

$\int x^3\cos h2xdx$.

Short answer

The value of$\int x^3\cos h2xdx$is,$\frac{\left(4x^3+6x\right)\sin h\left(2x\right)+\left(-6x^2-3\right)\cos h\left(2x\right)}{8}+C$.

Step-by-step solution

Step 1. Given information

$\int x^3\cos h2xdx$.

Step 2. Now evaluate the given integral.

Integrating by parts, $\int f{g}^{\text{'}}=fg-\int f^{\text{'}}g$.

Here, $f=x^3,g^{\text{'}}=\cos h\left(2x\right).$

$$\begin{gathered} f^{\text{'}}=3x^2,g=\frac{\sin h\left(2x\right)}{2} \\ \int x^3\cos h\left(2x\right)dx=x^3\frac{\sin h\left(2x\right)}{2}-\int 3x^2\frac{\sin h\left(2x\right)}{2}dx \\ =x^3\frac{\sin h\left(2x\right)}{2}-\left[\frac{3}{2}\int x^2\sin h\left(2x\right)dx\right] \\ =x^3\frac{\sin h\left(2x\right)}{2}-\left[\frac{3}{2}\left[\frac{x^2\cos h\left(2x\right)}{2}-\int x\cos h\left(2x\right)dx\right]\right] \\ =x^3\frac{\sin h\left(2x\right)}{2}-\frac{3}{4}{x}^2\cos h\left(2x\right)+\frac{3}{2}\int x\cos h\left(2x\right)dx \end{gathered}$$

Step 3. Use the Integration by parts method in the required places in the problem.

$$\begin{gathered} \int x^3\cos h\left(2x\right)dx=x^3\frac{\sin h\left(2x\right)}{2}-\frac{3}{4}{x}^2\cos h\left(2x\right)+\frac{3}{2}\left[x\frac{\sin h\left(2x\right)}{2}-\int \frac{\sin h\left(2x\right)}{2}dx\right] \\ =x^3\frac{\sin h\left(2x\right)}{2}-\frac{3}{4}{x}^2\cos h\left(2x\right)+\frac{3}{2}\frac{x\sin h\left(2x\right)}{2}-\frac{3}{2}\int \frac{\sin h\left(2x\right)}{2}dx \\ =x^3\frac{\sin h\left(2x\right)}{2}-\frac{3}{4}{x}^2\cos h\left(2x\right)+\frac{3}{2}\frac{x\sin h\left(2x\right)}{2}-\frac{3}{2}\frac{\cos h\left(2x\right)}{4} \\ =\frac{2x^3\sin h\left(2x\right)+3x\sin h\left(2x\right)}{4}+\left(\frac{-6x^2\cos h\left(2x\right)-3\cos h\left(2x\right)}{8}\right) \\ =\frac{\left(4x^3+6x\right)\sin h\left(2x\right)}{8}+\left(\frac{\left(-6x^2-3\right)\cos h\left(2x\right)}{8}\right) \\ =\frac{\left(4x^3+6x\right)\sin h\left(2x\right)+\left(-6x^2-3\right)\cos h\left(2x\right)}{8}+C \end{gathered}$$

Therefore, the required value is, $\frac{\left(4x^3+6x\right)\sin h\left(2x\right)+\left(-6x^2-3\right)\cos h\left(2x\right)}{8}+C$.