How do you find the antiderivative of cos2(x) ?

Question

Samantha Brown · Accepted Answer

The secret to figuring out this integral is to use an identity, in this case, the cosine double-angle identity.
Since $\cos (2 x) = \cos^{2} (x) - \sin^{2} (x)$ . We can rewrite this using the Pythagorean Identity to say that $\cos (2 x) = 2 \cos^{2} (x) - 1$ . Solving this for $\cos^{2} (x)$ shows us that $\cos^{2} (x) = \frac{\cos (2 x) + 1}{2}$
Thus:
$\int \cos^{2} (x) d x = \frac{12}{\int} \cos (2 x) + 1 d x$
Now that we have separated this, we can locate the antiderivative.
$= \frac{12}{\int} \cos (2 x) d x + \frac{12}{\in} 1 d x$
$= \frac{14}{\int} 2 \cos (2 x) d x + \frac{12}{x}$
$= \frac{14}{\sin (2 x)} + \frac{12}{x} + C$

ol3i4c5s4hr · Accepted Answer

We can&#039;t just integrate cos2(x) as it is, so we want to change it into another form, which we can easily do using trig identities.Recall the double angle formula: cos⁡(2x)=cos2(x)−sin2(x). We also know the trig identity sin2(x)+cos2(x)=1, so combining these we get the equation cos⁡(2x)=2cos2(x)−1.Now we can rearrange this to give: cos2(x)=1+cos⁡(2x)2So we have an equation which gives cos2(x) in a nicer form which we can easily integrate using the reverse chain rule.This eventually gives us an answer of x2+sin⁡(2x)4+C

nick1337 · Accepted Answer

Th antiderivative is pretty much the same as the integral, except it;s more general,
so I'll do the indefinite integral.
$\begin{matrix} \cos^{2} d d x \\ An identify for \cos^{2} x is: \\ \cos^{2} x = \frac{1 + \cos (2 x)}{2} \\ \Rightarrow \frac{1}{2} \int 1 + \cos (2 x) d x \\ Since \frac{d}{d x} [\sin (2 x)] = 2 (2 x) \int \cos (2 x) d x = \frac{1}{2} \sin (2 x) \\ \sin (2 x) = 2 \sin x \cos x so \frac{1}{2} \sin (2 x) = \sin x \cos x \\ \Rightarrow \frac{1}{2} [x + \frac{1}{2} \sin (2 x)] + C \\ = \frac{x}{2} + \frac{1}{4} \sin (2 x) + C \\ = \frac{x}{2} + \frac{1}{2} \sin x \cos x + C \end{matrix}$

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

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