Professor Dave here, let’s integrate trig functions.
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 are sine, cosine, tangent, cosecant, secant, and cotangent.
If you’ve forgotten what these are, go check out my trig tutorials to review, otherwise
let’s get to integrating.
The good news is that there are a few things that we can simply memorize when it comes
to integrating trig functions, and these will become ingrained over time and with practice.
Let’s start with cosine x.
Earlier, when we discussed differentiation of trig functions, we learned that the derivative
of sine is cosine.
If this doesn’t sound familiar, go and check out that tutorial now, because all of the
derivatives of trig functions that we learned before will be used here, just in reverse.
In other words, if the derivative of sine is cosine, then the integral of cosine must be sine.
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 integral
of cosine x is sine x plus C.
We also know that the derivative of cosine is negative sine.
So if we need to take the integral of sine x, let’s turn it into a form that lists
a common derivative, simply by writing negative the quantity negative sine x.
Then we can pull this first negative outside of the integral, since negative one is a constant,
and then integrating negative sine gives us cosine.
So the integral of sine x is negative cosine x plus C. The derivative of tangent is secant
squared, so if we have the integral of secant squared x, that will be tangent x plus C.
The derivative of cotangent is negative cosecant squared, so if we see the integral of cosecant
squared x, again we make this the double negative, bring one out of the integral, and we end
up with negative cotangent x plus C. Then there were the two weirder ones.
The derivative of secant is secant tangent, so the integral of secant x tangent x will
be secant x plus C. And then lastly, the derivative of cosecant is negative cosecant cotangent,
so if we see the integral of cosecant x cotangent x, we do the double negative thing one more
time, bring one of them outside, and we end up with negative cosecant x plus 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 to have to recognize these common derivatives
when we see them, and then to integrate, we just go backwards to the original function.
Let’s put this to use.
For example, how about the integral of ten x to the fourth minus two secant squared x.
Let’s make things easy on ourselves and split this difference up into separate integrals.
The first one is no problem, ten x to the fourth becomes ten x to the fifth over five,
or two x to the fifth.
Then for the other one, we can pull the two out here so that we aren’t distracted, and
we can recall that secant squared is the derivative of tangent, so this can become minus two tangent
x, and then finally we finish with plus C. Not too bad, right?
Of course, sometimes the common derivative won’t just be sitting there for you to spot.
You may have to manipulate things a little bit.
How about the integral of cosine x over sine squared x.
Again, there is no product rule or quotient rule for integrating, so we can’t do anything with this.
But when in doubt, just rewrite things in another form, and see what happens.
What if we write out sine squared as sine times sine.
We could then split this into the product of two fractions, one over sine times cosine over sine.
Then we could take one over sine and turn that into cosecant, and we can also take cosine
over sine and turn that into cotangent.
Now we have cosecant x cotangent x, and that is one of the common derivatives we have memorized,
so this will become negative cosecant x plus C. Let’s do just one more.
How about sine squared x plus cosine squared x, over cosine squared x.
Here, we are going to have to use one of our trig identities to change things around.
Well we know that sine squared plus cosine squared equals one, so let’s make this one
over cosine squared, which we know is the same as secant squared.
The integral of secant squared is tangent, so we get tangent x plus C.
If you didn’t spot that, don’t worry.
Even if you split this up into fractions to start, getting sine squared over cosine squared
plus cosine squared over cosine squared, that will simplify to tangent squared plus one.
And another one of the Pythagorean identities tells us that tangent squared plus one equals
secant squared, so we would have ended up at the same answer anyway, assuming that we
have our basic trig identities memorized.
And if we find trig functions inside an integrand that is a sum or difference, that won’t
change things at all, we will just have to get used to splitting them up and applying
the different rules we know depending on which term we are looking at.
Let’s check comprehension.