Joan Thompson

2021-12-29

Use Table of Integrals to evaluate the integral. ${x}^{3}\mathrm{sin}\left({x}^{2}+10\right)dx$.

Cheryl King

$\int {x}^{3}\mathrm{sin}\left({x}^{2}+10\right)dx$
Let ${x}^{2}+10=t$
2xdx=dt
$\frac{1}{2}\int \left(t-10\right)\mathrm{sin}tdt$
$=\frac{1}{2}\left[\int t\cdot \mathrm{sin}tdt-10\int \mathrm{sin}tdt\right]$
$=\frac{1}{2}\left[\left(\mathrm{sin}t-t\mathrm{cos}t+c\right)-10\left(-\mathrm{cos}t\right)\right]$
$=\frac{1}{2}\left[\mathrm{sin}t-t\mathrm{cos}t+10\mathrm{cos}t\right]+C$
$=\frac{1}{2}\left[\mathrm{sin}\left({x}^{2}+10\right)+10\mathrm{cos}\left({x}^{2}+10\right)-\left({x}^{2}+10\right)\mathrm{cos}\left({x}^{2}+10\right)\right]+C$

hysgubwyri3

Let's place the expression 2 * x under the differential's sign, i.e.

The initial integral can then be expressed as follows: The

formula for integration by parts:
$\int U\left(x\right)\cdot dV\left(x\right)=U\left(x\right)\cdot V\left(x\right)-\int V\left(x\right)\cdot dU\left(x\right)$
Put
U=x

Then:
dU=dx
$V=-\frac{\mathrm{cos}\left(x+10\right)}{2}$
Thus:

Find the integral

Result:
$x\cdot \frac{\mathrm{sin}\left(x+10\right)}{2}=-\frac{x\cdot \mathrm{cos}\left(x+10\right)}{2}+\frac{\mathrm{sin}\left(x+10\right)}{2}+C$
To write down the final answer, it remains to substitute ${x}^{2}$ instead of t.
$-{x}^{2}\cdot \frac{\mathrm{cos}\left({x}^{2}+10\right)}{2}+\frac{\mathrm{sin}\left({x}^{2}+10\right)}{2}+C$

Vasquez

$\begin{array}{}\int {x}^{3}×\mathrm{sin}\left({x}^{2}+10\right)dx\\ \int \frac{t×\mathrm{sin}\left(t\right)-10\mathrm{sin}\left(t\right)}{2}dt\\ \frac{1}{2}×\int t×\mathrm{sin}\left(t\right)-10\mathrm{sin}\left(t\right)dt\\ \frac{1}{2}×\left(\int t×\mathrm{sin}\left(t\right)dt-\int 10\mathrm{sin}\left(t\right)dt\right)\\ \frac{1}{2}×\left(-t×\mathrm{cos}\left(t\right)+\mathrm{sin}\left(t\right)+10\mathrm{cos}\left(t\right)\right)\\ \frac{1}{2}×\left(-\left({x}^{2}+10\right)×\mathrm{cos}\left({x}^{2}+10\right)+\mathrm{sin}\left({x}^{2}+10\right)+10\mathrm{cos}\left({x}^{2}+10\right)\right)\\ \frac{\mathrm{cos}\left({x}^{2}+10\right)×\left(-{x}^{2}-10\right)+\mathrm{sin}\left({x}^{2}+10\right)}{2}+5\mathrm{cos}\left({x}^{2}+10\right)\\ Solution:\\ \frac{\mathrm{cos}\left({x}^{2}+10\right)×\left(-{x}^{2}-10\right)+\mathrm{sin}\left({x}^{2}+10\right)}{2}+5\mathrm{cos}\left({x}^{2}+10\right)+C\end{array}$