Marla Payton

Answered

2021-12-26

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

Answer & Explanation

Samantha Brown

Expert

2021-12-27Added 35 answers

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:

$\int {\mathrm{cos}}^{2}\left(x\right)dx=\frac{12}{\int}\mathrm{cos}\left(2x\right)+1dx$

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

$=\frac{12}{\int}\mathrm{cos}\left(2x\right)dx+\frac{12}{\in}1dx$

$=\frac{14}{\int}2\mathrm{cos}\left(2x\right)dx+\frac{12}{x}$

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

ol3i4c5s4hr

Expert

2021-12-28Added 48 answers

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$\frac{x}{2}+\frac{\mathrm{sin}\left(2x\right)}{4}+C$

Recall the double angle formula:

Now we can rearrange this to give:

So we have an equation which gives

This eventually gives us an answer of

nick1337

Expert

2022-01-08Added 573 answers

Th antiderivative is pretty much the same as the integral, except it;s more general,

so I'll do the indefinite integral.

$\begin{array}{}{\mathrm{cos}}^{2}ddx\\ \text{An identify for}\text{}{\mathrm{cos}}^{2}x\text{}\text{is:}\\ {\mathrm{cos}}^{2}x=\frac{1+\mathrm{cos}(2x)}{2}\\ \Rightarrow \frac{1}{2}\int 1+\mathrm{cos}(2x)dx\\ \text{Since}\text{}\frac{d}{dx}[\mathrm{sin}(2x)]=2(2x)\text{}\int \mathrm{cos}(2x)dx=\frac{1}{2}\mathrm{sin}(2x)\\ \mathrm{sin}(2x)=2\mathrm{sin}x\mathrm{cos}x\text{so}\frac{1}{2}\mathrm{sin}(2x)=\mathrm{sin}x\mathrm{cos}x\\ \Rightarrow \frac{1}{2}[x+\frac{1}{2}\mathrm{sin}(2x)]+C\\ =\frac{x}{2}+\frac{1}{4}\mathrm{sin}(2x)+C\\ =\frac{x}{2}+\frac{1}{2}\mathrm{sin}x\mathrm{cos}x+C\end{array}$

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