Concepcion Hale

2022-01-03

Evaluate integral :
$\int {\mathrm{cos}}^{4}\left(x\right)dx$

Bertha Jordan

${\mathrm{cos}}^{4}\left(x\right)={\left(\frac{1+\mathrm{cos}\left(2x\right)}{2}\right)}^{2}=\frac{1+{\mathrm{cos}}^{2}\left(2x\right)+2\mathrm{cos}\left(2x\right)}{4}$
$=\frac{1+\frac{1+\mathrm{cos}\left(4x\right)}{2}+2\mathrm{cos}\left(2x\right)}{4}$
which gives us
${\mathrm{cos}}^{4}\left(x\right)=\frac{3+4\mathrm{cos}\left(2x\right)+\mathrm{cos}\left(4x\right)}{8}$
Now you should be able to integrate this off.

jean2098

Using the reduction formulae,
$\int {\mathrm{cos}}^{n}xdx=\frac{{\mathrm{cos}}^{n-1}x\mathrm{sin}x}{n}+\frac{n-1}{n}\int {\mathrm{cos}}^{n-2}xdx$
Putting $n=2$
$\int {\mathrm{cos}}^{2}xdx=\frac{\mathrm{cos}x\mathrm{sin}x}{2}+\frac{1}{2}\int dx=\frac{\mathrm{cos}x\mathrm{sin}x}{2}+\frac{1}{2}x+C$
Putting $n=4$,
$\int {\mathrm{cos}}^{4}xdx=\frac{{\mathrm{cos}}^{3}x\mathrm{sin}x}{4}+\frac{3}{4}\int {\mathrm{cos}}^{2}xdx$

Vasquez

${\mathrm{cos}}^{4}x={\mathrm{cos}}^{2}x-{\mathrm{cos}}^{2}x{\mathrm{sin}}^{2}x⇒$
$⇒I:=\int {\mathrm{cos}}^{4}xdx=\int {\mathrm{cos}}^{2}xdx-\int {\mathrm{cos}}^{2}x{\mathrm{sin}}^{2}xdx$
$=\frac{x+\mathrm{cos}x\mathrm{sin}x}{2}+\int \mathrm{sin}x{\mathrm{cos}}^{2}x\left(-\mathrm{cos}{\right)}^{\prime }xdx$
Now by parts in the last integegral:

so
$I:=\frac{x\mathrm{cos}x\mathrm{sin}x}{2}-\frac{1}{3}{\mathrm{cos}}^{3}x\mathrm{sin}x+\frac{1}{3}I⇒...$

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