How do you find the antiderivative of ${cos}^{2} (x)$ $cos^2 (x)$ ?

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How do you find the antiderivative of ${cos}^{2} (x)$ $cos^2 (x)$ ?

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Answer 1 · 2016-07-31T14:56:55

$\frac{1}{4} sin (2 x) + \frac{1}{2} x + C$ $1/4sin(2x)+1/2x+C$

Explanation:

The trick to finding this integral is using an identity--here, specifically, the cosine double-angle identity.

Since $cos (2 x) = {cos}^{2} (x) - {sin}^{2} (x)$ $cos(2x)=cos^2(x)-sin^2(x)$ , we can rewrite this using the Pythagorean Identity to say that $cos (2 x) = 2 {cos}^{2} (x) - 1$ $cos(2x)=2cos^2(x)-1$ . Solving this for ${cos}^{2} (x)$ $cos^2(x)$ shows us that ${cos}^{2} (x) = \frac{cos (2 x) + 1}{2}$ $cos^2(x)=(cos(2x)+1)/2$ .

Thus:

$\int {cos}^{2} (x) d x = \frac{1}{2} \int cos (2 x) + 1 d x$ $intcos^2(x)dx=1/2intcos(2x)+1dx$

We can now split this up and find the antiderivative.

$= \frac{1}{2} \int cos (2 x) d x + \frac{1}{2} \int 1 d x$ $=1/2intcos(2x)dx+1/2int1dx$

$= \frac{1}{4} \int 2 cos (2 x) d x + \frac{1}{2} x$ $=1/4int2cos(2x)dx+1/2x$

$= \frac{1}{4} sin (2 x) + \frac{1}{2} x + C$ $=1/4sin(2x)+1/2x+C$

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