We now understand what indefinite integrals are and we can evaluate them for simple polynomial functions. But as we know there are many types of functions that are more complicated than this and we have to be able to integrate them too.
So let's start with the trig functions that we learned in trigonometry.
These Trigonometric functions are:
- sine
- cosine
- tangent
- cosecant
- secant and
- cotangent
Integral of Trig Functions
Let's get trigonometric integrals.
The good news is that there are a few things that we can simply memorize when we have to take the integral of trig functions and these will become ingrained over time and with practice.
Integration of Cos x:
Let's start with cos x. Earlier when we discussed differentiation of trig functions we learned that the derivative of sinx is cosx.
$$ (sin x)' = cos x $$
In other words if the derivative of sinx is cosx then the integration of cos x must be sinx.
$$ \int cos x \; dx \;=\; sin x $$
Whatever happens one way, the opposite thing happens when we go the other way and of course as with any indefinite integral, we must add the constant "c". So the integration of cos x is sinx plus c.
$$ \int cos x \; dx \;=\; sin x \;+\; c $$
Integration of Sinx:
We also know that the derivative of cosx is negative sinx.
$$ (cos x)' = -sin x $$
So if we need to take the integral of sinx, let's turn it into a form that lists a common derivative, simply by writing negative the quantity negative sinx, then we can pull the first negative outside of the integral, since negative 1 is a constant.
$$ - \int - sin x \; dx $$
and then integrating negative sinx gives us cosx.
$$ \int sin x \; dx \;=\; -cosx \;+\; c $$
So the integration of sinx is negative cosx plus C.
Integral of Secx
The derivative of tanx is secant squared x.
$$ (tan x)' = sec^2x $$
So if we have the integral of sec squared x that will be tanx + c.
$$ \int sec^2x \; dx \;=\; tanx + c $$
Integral of Cscx
The derivative of cotx is negative cosecant squared.
$$ (cot x)' = -csc^2x $$
So if we see the integral of cosecant squared X, again just like the integral of sinx, we make the double negative and bring one out of the integral.
$$ - \int - csc^2x \; dx $$
And now we end up integrating cosec x with negative cot X plus C.
$$ - \int - csc^2x \; dx \;=\; -cot x \;+\; c $$
Integral of secx tanx
Then there were the two weirder ones. The derivative of secx is secx tanx.
$$ (sec x)' = sec x \; tan x $$
So the integral of secx tanx will be secx plus C
$$ \int secx \; tanx \; dx \;=\; sec x \;+\; c $$
Integral of Cscx Cotx
And now the derivative of cosecant is negative cosecant cotangent.
$$ (csc x)' = -csc x \; cot x $$
So if we see the integral of cscx cotx, we do the double negative thing one more time just like the integration of sinx and csc2 x and bring one of them outside.
$$ - \int - csc x \; cot x \; dx $$
And we end up with negative cosecant X plus C.
$$ - \int - csc x \; cot x \; dx \;=\; + \; c $$
So the good news is that if we have the derivatives of the six trig functions memorized we don't really have to learn anything new.
We just have to recognize the common derivatives when we see them and then to integrate we just go backwards to the original function. In this way we will be able to get integral of trig functions/trigonometric integrals.