# Can we calculate integral of cos^4x as (cos^2x)^2 instead of cos^3x*cosx?

Jun 13, 2018

$\int {\cos}^{4} x \mathrm{dx} = \frac{3 x}{8} + \sin \frac{2 x}{4} + \sin \frac{4 x}{32} + C$

#### Explanation:

Use the trigonometric identity:

${\cos}^{2} \alpha = \frac{1 + \cos \left(2 \alpha\right)}{2}$

Then:

$\int {\cos}^{4} x \mathrm{dx} = \int {\left({\cos}^{2} x\right)}^{2} \mathrm{dx}$

$\int {\cos}^{4} x \mathrm{dx} = \int {\left(1 + \cos \left(2 x\right)\right)}^{2} / 4 \mathrm{dx}$

using the linearity of the integral:

$\int {\cos}^{4} x \mathrm{dx} = \frac{1}{4} \int \mathrm{dx} + \frac{1}{2} \int \cos \left(2 x\right) \mathrm{dx} + \frac{1}{4} \int {\cos}^{2} \left(2 x\right) \mathrm{dx}$

$\int {\cos}^{4} x \mathrm{dx} = \frac{x}{4} + \sin \frac{2 x}{4} + \frac{1}{4} \int \frac{1 + \cos \left(4 x\right)}{2} \mathrm{dx}$

$\int {\cos}^{4} x \mathrm{dx} = \frac{x}{4} + \sin \frac{2 x}{4} + \frac{1}{8} \int \mathrm{dx} + \frac{1}{8} \int \cos \left(4 x\right) \mathrm{dx}$

$\int {\cos}^{4} x \mathrm{dx} = \frac{3 x}{8} + \sin \frac{2 x}{4} + \sin \frac{4 x}{32} + C$