The formula of the sine integral
\displaystyle \int \sin u \cdot du = - \cos u
Let’s see some examples for integrals of sin x:
Example 1. Integral of sin 2x
\displaystyle \int \sin(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 replaced the terms of x for u:
\displaystyle \int \sin(u) \cfrac{du}{2}
By properties of integrals we extract the \frac{1}{2} from the integral:
\displaystyle \cfrac{1}{2}\int\sin(u) \ du
Now we apply directly the formula of the sine integral:
\displaystyle \cfrac{1}{2}\int \sin (u) \ du = \left (\cfrac{1}{2}\right)(-\cos(u))
And finally we substitute the u for x and the answer will be:
-\cfrac{1}{2}\cos(2x)
Example 2. Integral of sin2 x
\displaystyle \int \sin^{2}(x) \ dx =
The fastest way to do this integral is to review the formula in the integrals form and you’re done. Another way is the following:
For the resolution of this integral, we need to remember the following trigonometric identity:
\sin^{2}(x) = \cfrac{1}{2} - \cfrac{1}{2} \cos(2x)
Substituting the \sin^{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 done, it would 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 the x for u 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 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)
Thanks for being here at this time : )
