The integral cosine formula is:
\displaystyle \int \cos u \cdot du = \sin u
Let’s see some examples for cosine integrals.
Example 1. Integral of cos2x
\displaystyle \int \cos(2x) \ dx=
We substitute the 2x for u, we derive and we pass dividing the 2:
u = 2x \quad \Rightarrow \quad du = 2 \ dx \quad \Rightarrow \quad \cfrac{du}{2} = dx
And we replace the terms for u:
\displaystyle \int \cos(u) \cfrac{du}{2}
By properties of integrals we extract the \frac{1}{2} from the integral:
\displaystyle \cfrac{1}{2}\int\cos(u) \ du
Now we apply directly the formula of the cosine integral:
\displaystyle \cfrac{1}{2} \int \cos (u) \ du = \left (\cfrac{1}{2}\right)(\sin(u))
And finally we substitute the u for 2x and the answer will be:
\cfrac{1}{2}\sin(2x)
Example 2. Integral of square cosine
\displaystyle \int \cos^{2}(x) \ dx =
The fastest way to do this integral is to review the formula in the Integrals Form and that’s it. Another way is the following:
For the resolution of this integral, we need to remember the following trigonometric identity:
\cos^{2}(x) = \cfrac{1}{2} + \cfrac{1}{2} \cos(2x)
Substituting the \cos^{2}(x) for \frac{1}{2}+\frac{1}{2}\cos(2x), we will have the following integral of a sum:
\displaystyle \int \left( \cfrac{1}{2} + \cfrac{1}{2} \cos(2x) \right) dx
By properties of integrals we will have a sum of integrals:
\displaystyle \int \cfrac{1}{2}\ dx + \int \cfrac{1}{2} \cos(2x) \ dx
The first integral can easily be made, it will be as follows:
\displaystyle \cfrac{1}{2} x + \int \cfrac{1}{2} \cos(2x) \ dx
We apply properties of the integrals to extract the \frac{1}{2} from the integral:
\displaystyle \cfrac{1}{2}x + \cfrac{1}{2} \int \cos(2x) \ dx
To solve the second integral, we have to do some procedures that were done in example 1, we substitute 2x for u, we derive and we pass dividing the 2:
u = 2x \quad \Rightarrow \quad du = 2 \ dx \quad \Rightarrow \quad \cfrac{du}{2} = dx
Now we substitute u‘s in the second integral with which we are working:
\displaystyle \cfrac{1}{2}x + \cfrac{1}{2} \int \cos(u) \cfrac{du}{2}
We take out the denominator 2 that is in du:
\displaystyle \cfrac{1}{2}x + \cfrac{1}{2}\cfrac{1}{2} \int \cos(u) \ du
We multiply the fractions of \frac{1}{2}:
\displaystyle \cfrac{1}{2} x + \cfrac{1}{4} \int \cos(u) \ du
We apply the cosine integration formula which is the following:
\displaystyle \int \cos(u) \ du = \sin (u)
Then the integration will remain as follows:
\cfrac{1}{2}x + \cfrac{1}{4} \sin(u)
And finally we substitute the u for 2x to obtain our result:
\cfrac{1}{2}x + \cfrac{1}{4}\sin(2x)
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