prsategazd

2021-12-11

Evaluate the integral.
$\int {\mathrm{cos}}^{2}x\mathrm{sin}\left(2x\right)dx$

temzej9

Step 1
Given integral to evaluate
$\int {\mathrm{cos}}^{2}x\mathrm{sin}\left(2x\right)dx$
substitute $\mathrm{sin}\left(2x\right)=2\mathrm{sin}x\mathrm{cos}x$
$\int {\mathrm{cos}}^{2}x\mathrm{sin}\left(2x\right)dx=\int {\mathrm{cos}}^{2}x2\mathrm{sin}x\mathrm{cos}xdx$
$=\int 2{\mathrm{cos}}^{3}x\mathrm{sin}xdx$
Step 2
Now put
$t=\mathrm{cos}x$
$dt=-\mathrm{sin}xdx$
so the integral becomes
$\int -2{t}^{3}dt=-2\int {t}^{3}dt$
$=-2\left[\frac{{t}^{4}}{4}\right]+c$
$=-\frac{{t}^{4}}{2}+c$
put back value of t.
$\int {\mathrm{cos}}^{2}x\mathrm{sin}\left(2x\right)dx=-\frac{{\mathrm{cos}}^{4}x}{2}+c$

ambarakaq8

$\int {\mathrm{cos}}^{2}\left(x\right)\mathrm{sin}\left(2x\right)dx$
$=\int -\left(-2{\mathrm{cos}}^{3}\left(x\right)\right)\mathrm{sin}\left(x\right)dx$
$=-2\int {u}^{3}du$
Now we calculate:
$\int {u}^{3}du$
$=\frac{{u}^{4}}{4}$
We substitute the already calculated integrals:
$-2\int {u}^{3}du$
$=-\frac{{u}^{4}}{2}$
Reverse replacement $u=\mathrm{cos}\left(x\right):$
$=-\frac{{\mathrm{cos}}^{4}\left(x\right)}{2}$
$=-\frac{{\mathrm{cos}}^{4}\left(x\right)}{2}+C$