lugreget9

2021-12-30

Evaluate the indefinite integral.
$\int \mathrm{cos}x\left(3\mathrm{sin}x-1\right)dx$

### Answer & Explanation

scomparve5j

Step 1
Consider the provided integral,
$\int \mathrm{cos}x\left(3\mathrm{sin}x-1\right)dx$
Evaluate the indefinite integral.
Apply the substitution method,
Let, $u=3\mathrm{sin}x-1⇒du=3\mathrm{cos}xdx$.
Step 2
Therefore the integral becomes,
$\int \mathrm{cos}x\left(3\mathrm{sin}x-1\right)dx=\int \frac{u}{3}du$
$=\frac{1}{3}\cdot \int udu$
$=\frac{1}{3}\cdot \frac{{u}^{1+1}}{1+1}+C$
$=\frac{{u}^{2}}{6}+C$
Step 3
Substitute back $u=3\mathrm{sin}x-1$ in the above integral,
$\int \mathrm{cos}x\left(3\mathrm{sin}x-1\right)dx=\frac{1}{6}{\left(3\mathrm{sin}\left(x\right)-1\right)}^{2}+C$
Hence.

Cheryl King

$\int \mathrm{cos}\left(x\right)\left(3\mathrm{sin}\left(x\right)-1\right)dx$
$=\frac{1}{3}\int udu$
$\int udu$
$=\frac{{u}^{2}}{2}$
$\frac{1}{3}\int udu$
$=\frac{{u}^{2}}{6}$
$=\frac{{\left(3\mathrm{sin}\left(x\right)-1\right)}^{2}}{6}$
$\int \mathrm{cos}\left(x\right)\left(3\mathrm{sin}\left(x\right)-1\right)dx$
$=\frac{{\left(3\mathrm{sin}\left(x\right)-1\right)}^{2}}{6}+C$

karton

$\int \mathrm{cos}\left(x\right)×\left(3\mathrm{sin}\left(x\right)-1\right)dx$
Remove the parentheses
$\int 3\mathrm{cos}\left(x\right)\mathrm{sin}\left(x\right)-\mathrm{cos}\left(x\right)dx$
Simplify the expression
$\int \frac{3}{2}×\mathrm{sin}\left(2x\right)-\mathrm{cos}\left(x\right)dx$
Calculate
$\int \frac{3\mathrm{sin}\left(2x\right)}{2}-\mathrm{cos}\left(x\right)dx$
Use properties of integrals
$\int \frac{3\mathrm{sin}\left(2x\right)}{2}dx-\int \mathrm{cos}\left(x\right)dx$
Evaluate the integrals
$-\frac{3\mathrm{cos}\left(2x\right)}{4}-\mathrm{sin}\left(x\right)$
$-\frac{3\mathrm{cos}\left(2x\right)}{4}-\mathrm{sin}\left(x\right)+C$