The integral formula of tan is:
\displaystyle \int \tan u \cdot du = \ln |\sec u| = -\log \cos u
Of so many trigonometric integrals, we will see some examples for tangent integrals.
Example 1. Integral of tan 2x
\displaystyle \int \tan(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 \tan(u) \cfrac{du}{2}
By properties of integrals we extract the \frac{1}{2} from the integral:
\displaystyle \cfrac{1}{2}\int\tan(u) \ du
Now we apply directly the tangent integral formula:
\displaystyle \cfrac{1}{2}\int \tan (u) \ du = \left (\cfrac{1}{2}\right)(-\log \cos(u))
And finally we substitute the u for 2x and the answer will be:
-\cfrac{1}{2}\log[\cos(2x)]
Example 2. Integral of square tangent
\displaystyle \int \tan^{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 to shred the integral a bit and review the Integral Form anyway at some point, let’s start:
To begin with the resolution of this integral, the first thing we have to do is apply the following trigonometric identity:
\tan^{2}x + 1 = \sec^{2}x
Now what you have to do is clear \tan^{2}x:
\tan^{2}x = \sec^{2}x - 1
Substituting \tan^{2}x into the integral, we will obtain the integral of a subtraction:
\displaystyle \int\left ( \sec^{2}x - 1 \right) \ dx
Separate the integral to a sum of integrals:
\displaystyle \int \sec^{2}x \ dx + \int - 1 \ dx
Applying properties of the integrals we will remove the -1 from the integral:
\displaystyle \int \sec^{2}x \ dx - 1 \int \ dx
To solve the first integral, we will review the Integral Form that shows us an integral formula of \sec^{2}x, therefore, the first integral would be as follows:
\displaystyle \int\sec^{2}x \ dx- \int \ dx = \tan x - \int \ dx
Solving the second integral, the answer will be the following:
\tan x - x
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