Marla Payton

2021-12-26

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

Samantha Brown

The secret to figuring out this integral is to use an identity, in this case, the cosine double-angle identity.
Since $\mathrm{cos}\left(2x\right)={\mathrm{cos}}^{2}\left(x\right)-{\mathrm{sin}}^{2}\left(x\right)$. We can rewrite this using the Pythagorean Identity to say that $\mathrm{cos}\left(2x\right)=2{\mathrm{cos}}^{2}\left(x\right)-1$. Solving this for ${\mathrm{cos}}^{2}\left(x\right)$ shows us that ${\mathrm{cos}}^{2}\left(x\right)=\frac{\mathrm{cos}\left(2x\right)+1}{2}$
Thus:

Now that we have separated this, we can locate the antiderivative.

$=\frac{14}{\mathrm{sin}\left(2x\right)}+\frac{12}{x}+C$

ol3i4c5s4hr

We can't just integrate ${\mathrm{cos}}^{2}\left(x\right)$ 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: $\mathrm{cos}\left(2x\right)={\mathrm{cos}}^{2}\left(x\right)-{\mathrm{sin}}^{2}\left(x\right)$. We also know the trig identity ${\mathrm{sin}}^{2}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right)=1$, so combining these we get the equation $\mathrm{cos}\left(2x\right)=2{\mathrm{cos}}^{2}\left(x\right)-1$.
Now we can rearrange this to give: ${\mathrm{cos}}^{2}\left(x\right)=\frac{1+\mathrm{cos}\left(2x\right)}{2}$
So we have an equation which gives ${\mathrm{cos}}^{2}\left(x\right)$ in a nicer form which we can easily integrate using the reverse chain rule.
This eventually gives us an answer of $\frac{x}{2}+\frac{\mathrm{sin}\left(2x\right)}{4}+C$

nick1337

Th antiderivative is pretty much the same as the integral, except it;s more general,
so I'll do the indefinite integral.

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