Mary Hammonds

2021-12-27

Evaluate the following integrals:
$\int {\mathrm{cos}}^{2}x\mathrm{sin}xdx$

Pansdorfp6

Step 1
In some cases, an integral can be simplified to a standard integral with an appropriate substitution. For example, the integral $\int {\mathrm{sin}}^{2}x\mathrm{cos}xdx$ can be converted to $\int {u}^{2}du$ with the substitution $u=\mathrm{sin}x$.
For the given problem, use the integral $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1},n\ne -1$. Find an appropriate substitution for the given integral to simplify the integrand to a simpler integrand.
Step 2
Integral to be computed is $\int {\mathrm{cos}}^{2}x\mathrm{sin}xdx$. Use the substitution, $\mathrm{cos}x=u$. Differentiating this, gives $-\mathrm{sin}xdx=du$. Apply this substitution and integrate using information from step 1.
$\int {\mathrm{cos}}^{2}x\mathrm{sin}xdx=\int {u}^{2}\left(-du\right)$
$=-\int {u}^{2}du$
$=-\frac{{u}^{3}}{3}+C$
$=-\frac{{\mathrm{cos}}^{3}x}{3}+C$
Hence, the integral is equal to $-\frac{{\mathrm{cos}}^{3}x}{3}+C$.

braodagxj

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

karton

$\int \mathrm{cos}\left(x{\right)}^{2}\mathrm{sin}\left(x\right)dx$
$\int -{t}^{2}dt$
$-\int {t}^{2}dt$
$-\frac{{t}^{3}}{3}$
$-\frac{\mathrm{cos}\left(x{\right)}^{3}}{3}$
$-\frac{\mathrm{cos}\left(x{\right)}^{3}}{3}+C$