# How do you find the antiderivative of cos^2 (x)?

Jul 30, 2016

$\int {\cos}^{2} \left(x\right) d x = \frac{x}{2} + \frac{\cos \left(x\right) \sin \left(x\right)}{2} + C$

#### Explanation:

int cos^2(x) d x=?

$\text{let us use the reduction formula :}$

${\cos}^{n} \left(x\right) d x = \frac{n - 1}{n} \int {\cos}^{n - 2} \left(x\right) d x + \frac{{\cos}^{n - 1} \left(x\right) \sin \left(x\right)}{n}$

$\text{Apply n=2}$

$\int {\cos}^{2} \left(x\right) d x = \frac{2 - 1}{2} \int {\cos}^{2 - 2} \left(x\right) d x + \frac{{\cos}^{2 - 1} \left(x\right) \sin \left(x\right)}{2}$

$\int {\cos}^{2} \left(x\right) d x = \frac{1}{2} \int {\cos}^{0} \left(x\right) d x + \frac{\cos \left(x\right) \sin \left(x\right)}{2}$

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

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

$\int {\cos}^{2} \left(x\right) d x = \frac{x}{2} + \frac{\cos \left(x\right) \sin \left(x\right)}{2} + C$

Remember that cos2x=cos^2x-sin^2x=>cos2x=2cos^2x-1=> cos^2x=1/2(1+cos2x)

Hence

int cos^2xdx=int 1/2 (1+cos2x)dx=1/2int 1dx+1/2intcos2xdx= 1/2x+1/4sin2x+c