# What is the integral of int sin^4(x) dx?

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#### Explanation

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#### Explanation:

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Dec 17, 2017

$\int \setminus {\sin}^{4} \left(x\right) \setminus \mathrm{dx} = \frac{3}{8} x - \frac{1}{4} \sin \left(2 x\right) + \frac{1}{32} \sin \left(4 x\right) + C$

#### Explanation:

This integral is mostly about clever rewriting of your functions. As a rule of thumb, if the power is even, we use the double angle formula. The double angle formula says:
${\sin}^{2} \left(\theta\right) = \frac{1}{2} \left(1 - \cos \left(2 \theta\right)\right)$

If we split up our integral like this,
$\int \setminus {\sin}^{2} \left(x\right) \cdot {\sin}^{2} \left(x\right) \setminus \mathrm{dx}$

We can use the double angle formula twice:
$\int \setminus \frac{1}{2} \left(1 - \cos \left(2 x\right)\right) \cdot \frac{1}{2} \left(1 - \cos \left(2 x\right)\right) \setminus \mathrm{dx}$

Both parts are the same, so we can just put it as a square:
$\int \setminus {\left(\frac{1}{2} \left(1 - \cos \left(2 x\right)\right)\right)}^{2} \setminus \mathrm{dx}$

Expanding, we get:
$\int \setminus \frac{1}{4} \left(1 - 2 \cos \left(2 x\right) + {\cos}^{2} \left(2 x\right)\right) \setminus \mathrm{dx}$

We can then use the other double angle formula
${\cos}^{2} \left(\theta\right) = \frac{1}{2} \left(1 + \cos \left(2 \theta\right)\right)$
to rewrite the last term as follows:
$\frac{1}{4} \int \setminus 1 - 2 \cos \left(2 x\right) + \frac{1}{2} \left(1 + \cos \left(4 x\right)\right) \setminus \mathrm{dx} =$

$= \frac{1}{4} \left(\int \setminus 1 \setminus \mathrm{dx} - \int \setminus 2 \cos \left(2 x\right) \setminus \mathrm{dx} + \frac{1}{2} \int \setminus 1 + \cos \left(4 x\right) \setminus \mathrm{dx}\right) =$

$= \frac{1}{4} \left(x - \int \setminus 2 \cos \left(2 x\right) \setminus \mathrm{dx} + \frac{1}{2} \left(x + \int \setminus \cos \left(4 x\right) \setminus \mathrm{dx}\right)\right)$

I will call the left integral in the parenthesis Integral 1, and the right on Integral 2.

Integral 1
$\int \setminus 2 \cos \left(2 x\right) \setminus \mathrm{dx}$

Looking at the integral, we have the derivative of the inside, $2$ outside of the function, and this should immediately ring a bell that you should use u-substitution.

If we let $u = 2 x$, the derivative becomes $2$, so we divide through by $2$ to integrate with respect to $u$:
$\int \setminus \frac{\cancel{2} \cos \left(u\right)}{\cancel{2}} \setminus \mathrm{du}$

$\int \setminus \cos \left(u\right) \setminus \mathrm{du} = \sin \left(u\right) = \sin \left(2 x\right)$

Integral 2
$\int \setminus \cos \left(4 x\right) \setminus \mathrm{dx}$

It's not as obvious here, but we can also use u-substitution here. We can let $u = 4 x$, and the derivative will be $4$:
$\frac{1}{4} \int \setminus \cos \left(u\right) \setminus \mathrm{dx} = \frac{1}{4} \sin \left(u\right) = \frac{1}{4} \sin \left(4 x\right)$

Completing the original integral
Now that we know Integral 1 and Integral 2, we can plug them back into our original expression to get the final answer:
$\frac{1}{4} \left(x - \sin \left(2 x\right) + \frac{1}{2} \left(x + \frac{1}{4} \sin \left(4 x\right)\right)\right) + C =$

$= \frac{1}{4} \left(x - \sin \left(2 x\right) + \frac{1}{2} x + \frac{1}{8} \sin \left(4 x\right)\right) + C =$

$= \frac{1}{4} x - \frac{1}{4} \sin \left(2 x\right) + \frac{1}{8} x + \frac{1}{32} \sin \left(4 x\right) + C =$

$= \frac{3}{8} x - \frac{1}{4} \sin \left(2 x\right) + \frac{1}{32} \sin \left(4 x\right) + C$

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