The sec integral formula is:
\displaystyle \int \sec u \cdot du = \log \left(\sec u + \tan u\right)
From so many trigonometric integrals, we will see some examples for the secant integral:
Example 1. Integral of the secant of x
\displaystyle \int \sec x \ dx
The secant integral can be deduced if you know some derivation formulas, then we will see how to get to the result of \log \left(\sec x + \tan x \right) with the formulas of the derivatives . Let’s begin!
To start with this integral, there is a very powerful trick that allows us to solve it, we will multiply the \sec x by a fraction that contains \sec x + \tan x in the denominator and in the numerator.
\displaystyle \int \sec x \cfrac{\sec x + \tan x}{\sec x + \tan x} dx
We will do the multiplication and the obtained result will be the following:
\displaystyle \int \cfrac{\sec^{2}x + \sec x \tan x}{\sec x + \tan x} dx
Then we will name u to \sec x+\tan x and we will derive it. Remember the derivation formula that say that the derivative of \sec x is equal to \sec x \tan x and the derivative of \tan x equals to \sec^{2}x:
u = \sec x + \tan x
du = \left(\sec x \tan x + \sec^{2} x\right) dx
Once we have u and du, we will substitute them in the integral, as follows:
\displaystyle \int \cfrac{1}{u} \ du
And now applying directly an integration formula, we will obtain the following result of our integral:
\displaystyle \int \cfrac{1}{u} \ du = \log (u)
Finally we substitute the u to obtain our final result:
\log\left(\sec x + \tan x \right)
Example 2. Integral of secant of 2x
\displaystyle \int \sec (2x) dx
The resolution process is very similar to the previous example, let’s start.
Let’s start replacing 2x for v, then we derive and clear:
dv = 2 \ dx
\cfrac{dv}{2} = dx
We substitute the integral 2x for v and dx for \frac{dv}{2}:
\displaystyle \int \sec v \cfrac{dv}{2}
Applying properties of the integrals we will take out the denominator 2 from the integral:
\displaystyle \cfrac{1}{2} \int \sec v \ dv
From here we can forget the \frac{1}{2} (obviously we will not forget the \frac{1}{2}) and do the integration directly with formula or apply the whole procedure already explained in the previous example.
Let’s multiply the \sec v by a fraction containing \sec v+\tan v in the numerator and in the denominator.
\displaystyle \cfrac{1}{2}\int \sec v \cfrac{\sec v + \tan v}{\sec v + \tan v} dv
We will do the multiplication and the obtained result will be the following:
\displaystyle \cfrac{1}{2} \int \cfrac{\sec^{2}v + \sec v \tan v}{\sec v + \tan v} dv
Then we will name u to \sec v+\tan v and we will derive it. Remember the derivation formula that say that the derivative of \sec x is equal to \sec x\tan x and the derivative of \tan x equals to \sec^{2} x:
u = \sec v + \tan v
du = \left(\sec v \tan v + \sec^{2} v\right) dv
Once we have u and du, we will substitute them in the integral, as follows:
\displaystyle \cfrac{1}{2} \int \cfrac{1}{u} \ du
And now applying directly an integration formula, we will obtain the following result of our integral:
\displaystyle \cfrac{1}{2} \int \cfrac{1}{u} \ du = \log (u)
We substitute the u to obtain:
\cfrac{1}{2} \log\left(\sec v + \tan v \right)
And finally we substitute v for 2x to obtain our final result:
\cfrac{1}{2} \log \left(\sec 2x + \tan 2x\right)
Example 3. Integral of square secant of x
\displaystyle \int \sec^{2}x \ dx
Maybe some of them are a bit conflicted about the resolution of this integral, but we know very well that integrals and derivatives are linked by a strong bond of friendship, which means that if someone derives something and the result of that something is integrated, the result is going to be that something with which it started. If someone derives x, he will get 1, if someone integrates 1, someone gets x.
The derivative of \tan x equals \sec^{2}x\ dx, that means that the integral of \sec^{2}x\ dx is \tan x, so our end result of integral \sec^{2}x will be:
\tan x
Example 4. Integral of secant cubed of x
\displaystyle \int \sec^{3}(x) dx
To solve this integral, what we need is to use the secant reduction formula, so by applying the formula, we obtain the following:
\displaystyle \cfrac{\sin x \sec^{2} x}{2} + \cfrac{1}{2} \int \sec x \ dx
Using the following identities …
\sec x = \cfrac{1}{\cos x}
\tan x = \cfrac{\sin x}{\cos x}
… we will arrive at our next expression:
\displaystyle = \cfrac{1}{2} \sec x \tan x + \cfrac{1}{2} \int \sec x \ dx
Applying the secant integral of x, we will reach our final result:
= \cfrac{1}{2} \sec x \tan x + \cfrac{1}{2} \ln \left| \sec x + \tan x \right|
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