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### Integration of Trigonometric Functions

Main formula:

Hints and tips:

• Remember to check which of the powers is odd. Then let u be the other one. (E.g. if you have cos5 x, then use u = sin x.)

• If both powers are even, use the half-angle formulas.

• If you have an even power of secant, use u = tan x. If you have an odd power of tangent, use u = sec x.

• If you have an odd power of secant and an even power of tangent, then follow this process:

1. Convert tangents to secants, two at a time, using tan² +1 = sec² . You’ll end up with odd powers of secant.

2. Use integration by parts twice (u = secn−2 x, dv = sec² x dx) to cut down from ∫ secn x dx to ∫ secn−2 x dx, so you eventually get back to ∫ sec x dx. Then remember the formula:

### Integration of Trigonometric Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Important Equation 0:07
• Powers (Odd and Even)
• What To Do
• Lecture Example 1 1:37
• Lecture Example 2 3:12
• Half-Angle Formulas 6:16
• Both Powers Even
• Lecture Example 3 7:06
• Lecture Example 4 10:59

### Transcription: Integration of Trigonometric Functions

OK, lset us try a new example.0000

We have here tan3(x) × sec(x) and remember that the trick here is to look at the powers.0004

You first look at the powers of tan(x).0009

Is that even?0012

Well 3 is odd so that does not work so we cannot use our strategy there.0014

The power on sec(x) is 1.0020

We ask, is that an odd number?0023

That does work, so we are going to use u = sec(x).0026

Then du = sec(x) tan(x) dx.0033

The point of that is that we can then convert this integral into something just involving sec(x).0043

We have tan3(x), so we will write that as tan2(x) × tan(x) × sec(x) dx.0051

Now the u is going to be sec(x), but we still need to have du.0064

sec(x) tan(x).0066

We have that right here, sec(x) tan(x).0069

The tan2(x), we need to convert that into secants.0073

Since we have an even number of tangents, we can do that.0078

Remember our old Pythagorean Identity, tan2(x) + 1 = sec2(x).0081

The tan2(x) converts into sec2(x) - 1.0090

tan(x) sec(x) dx.0099

Now we are set up to put our substitution in.0103

Sec2(x) - 1 is u2 - 1.0106

tan(x) sec(x) dx, that is just du.0112

All of a sudden we have a really easy integral.0117

We can integrate that quickly to be u3/3 - u.0121

Then we substitute back into sec3(x)/3 - sec(x) + C.0126

So, again, the reason that worked is because we were checking the odds and even powers on the tangent and the secant.0143

It worked because we had an odd power of sec(x).0155

There is one more example I want to work through with you.0000

It is a pretty tricky one.0003

It is sec3(x).0005

This one is not so obvious.0006

What we are going to do is let u = sec(x).0010

But the point of that is not to make a substitution.0017

The point of that is to use integration by parts.0022

If u = sec(x), then dv = sec2(x) dx.0024

That is the stuff left over after you use one of these secants.0032

That is not obvious that we should do it that way.0036

You might ask how did you know to break it up that way?0041

Well, here is the idea.0044

I know that sec2 is the derivative of tangent.0046

So, by making dv be sec2 v becomes tan(x).0050

That is a relatively simple integration.0056

By making u = sec(x), we have du = sec(x) tan(x) dx.0061

That looks a little more complicated.0070

We will see how it works out.0074

Remember the formula for integration by parts.0075

It is always uv - the integral of vdu.0077

So, uv is sec(x) tan(x) - the integral of vdu.0081

So vdu looks a little complicated.0086

We have sec(x) tan(x) dx × tan(x).0096

So sec(x) × tan2(x) dx.0102

This is looking a little complicated.0108

What I am going to do is take this tan(x) and remember the Pythagorean Identity.0114

tan2(x) + 1 is sec2(x).0116

I am going to write that tan(x) as sec2(x) - 1.0123

The whole thing becomes sec(x) tan(x) - now I will distribute this secant across these two terms.0135

So we have - the integral of sec3(x) dx.0147

+ the integral of sec(x) dx.0156

Now, it looks like we have taken what seemed like a pretty hard integral and replaced it by another hard integral.0163

And, an integral that is exactly the same as what we started with.0171

It looks like it is getting worse and worse.0175

Let me show you how you can resolve those two.0178

First of all, the sec(x) dx, there is a trick here.0181

Which is to multiply by sec(x) + tan(x)/sec(x) + tan(x).0184

You are multiplying by 1, not changing it.0204

But you get something very interesting when you multiply that into the integral.0209

The numerator will be sec(x) × (sec(x) + tan(x)).0213

So, sec2(x) + sec(x) tan(x).0220

The denominator is going to be sec(x) + tan(x) all multiplied by dx.0224

Look at this.0238

If you look at the denominator, sec(x) + tan(x), the derivative of sec(x) is sec(x) tan(x).0240

The derivative of tan(x) is sec2(x).0250

So the numerator is the derivative of the denominator.0255

So this last integral here is a derivative divided by the original function.0259

That last integral just integrates to ln(sec(x)) + tan(x).0266

That is a pretty special trick that is probably worth memorizing, just so you can do the integral of sec(x) dx.0267

Or else, it is just worth remembering that the integral of sec(x) dx is ln(sec(x) + tan(x)).0287

It is a strange and unusual enough integral that is worth memorizing for its own sake.0296

Let us go back to our original integral, which I am going to call i, just as we did when we did integration by parts twice.0302

What we have now is i has been resolved into sec(x) tan(x) - now we have i again, + that integral that we just did.0311

ln(sec(x)) + tan(x).0328

Separate that part out and so again just like we did when we had to do integration by parts twice.0334

We can move this i over and get 2i = sec(x) tan(x) + ln(sec(x) + tan(x).0343

Again, we can solve for the original integral just by dividing both sides by 2.0364

We get i = 1/2 of the quantity sec(x) tan(x) + ln(abs(sec(x) + tan(x).0369

As always, we have to add the constant at the end.0386

So, that is a fairly complicated answer.0397

That is a fairly specialized problem.0401

When you have a power of sec, an odd power of secant all by itself.0405

You want to use this integration by parts, sort of reduction formula,0407

To reduce it down to a lower power of secant.0416

If you keep reducing it using this formula, you eventually get down to just sec(x) by itself.0420

Remember we said that we will memorize that the integral of sec(x) = ln(sec(x) + tan(x).0429

That is the end of this lecture on integration of trigonometric functions.0437

Hello, this is educator.com and today we are going to talk about integration of trigonometric functions.0000

The prototypical examples of these integrals is you will have an integral and some power of sine and some power of cosine.0006

The important thing to focus on here is what those powers are.0015

In particular, which one is odd.0022

This is really a game of odds and evens.0029

You want one of those powers to be odd.0031

If one of them, if they are both even, then the integral gets more difficult and we will talk about that later in the lecture.0034

So right now the important thing is to find one of those numbers and to make it the odd or hope that it is odd.0041

If they are both odd, then it really does not matter, you can just pick one.0047

The one that is odd, what you do with it, is you are going to make a substitution and let U be the other one.0051

U is going to be the other one, and dU will be either + or - the one that is odd.0076

It seems a little strange so it would be easier to understand if we work through some examples.0083

You will see why this strategy works so well.0089

Let us take a look at some examples.0093

Here is the first example.0095

Cos(x) times sin4X, dX.0098

Of course, cos(x) is the same as cos(x) to the 1.0105

So, you look at those numbers 1 and 4, and you see the odd one, that is the odd number.0109

What we are going to do is let U be the other one.0114

The other one is sin(x) and the point of doing that is dU is then cosin(x) dX.0120

So this integral, the cosin, and the dX, become the dU.0128

Those two become the dU, sin4X just becomes U4.0136

All of a sudden you have a really easy integral, the integral of U4.0144

Well, that is just U5/5 + C.0149

Then you substitute back, and you get sin5X + C.0156

Just to reiterate there, what made that work was, we looked at which power was odd, and that was the power on cosine.0167

Then, we let u be the other one, sin(x)0177

The reason we did that is because we had that one power of cosine left over to be the dU.0182

So let us try that on a little more complicated example.0188

Cos4X sinU3X dX.0191

Again, we look at those two powers and the odd one is 3.0198

We are going to let U be the other one.0201

U is cos(x) this time, and so dU, well the derivative of cosin is sin(x) dX0205

What we can do now is we can take this sinU3 and we can split that up into a sin2X × sin(x) dX.0207

The point here is that we save that sin(x) dX to be the dU.0231

Well, we are going to have to attach a negative sign, but that sin(x) dX is going to be the dU.0236

Then this sin2, we can convert that into 1 - cosin2x.0243

That is why we really had to have this odd even pattern going on.0249

It is because we save one sin to be the dU, or actually negative dU, and then you can convert sines into cosines two at a time.0256

If we save one sine to be dU, that will leave an even number of sines left over.0266

We can covert those all into cosines.0273

Our integral, after we make all of those substitutions, cos4 becomes u4.0276

The sin2 becomes 1 - cos2, and then becomes 1 - u4.0285

Sin(x) dX is dU, or actually negative dU.0292

Then we can combine these into one big polynomial.0297

Actually, I can bring this negative sign back inside, and make it the integral of u4 × u2 - 1 dU.0304

Combine those into u6 - u4 dU.0310

Then, that is an easy integral because then we can just use the power rule, u7/7 - u5/5.0321

Finally, we can convert that back into u as cos(x).0330

So, this is cos7X/7 - cos5X/5, and of course we have to attach a constant there.0337

Again, the key thing here was recognizing those powers.0350

We see 4, we see 3, 3 is the odd one, so we want to let u be the other one.0357

The other one was cos(x).0365

The point of that was it gives you one extra sin(x) left over to be the dU.0367

Now, we know what to do if either one of the powers is odd.0373

What if you have both powers being even?0382

That makes it a trickier problem.0384

What we are going to use are the half angle formulas.0386

If both sine and cosine are even powers, then we can not use the previous strategy.0391

That odd strategy does not work.0398

We are going to use the half angle formulas.0401

Cos2(x) = 1 + cos(2x)/20403

Sin2X = 1 - cos(2x)/2.0407

The point of those is that those are easy to integrate.0411

Even if we have to multiply those together a few times, as we will see in the example, it is not too bad to integrate.0415

Let us try out an example of that.0422

Here we have sin2x cos2x dX.0425

We look at those powers, we look for the odd one, and oh no, they are both even.0430

So, we are going to use those half angle formulas.0436

Remember, sin2x is 1 - cos(2x) over 2.0438

Cos2x is 1 + cos(2x)/2.0446

Now I am going to pull the two halves out of the integral to get them away.0455

So now that is on the outside.0460

Now, we have the integral of 1 - cos(2x) × 1 + cos(2x).0462

I will multiply those two together and we will get 1 - cos2(2x).0469

Well, the 1 is going to be easy to integrate, so I am not going to be worried about that.0477

But now we have a cos2(2x).0481

How do we integrate that?0484

Well, again, that is an even power of cosine.0486

We can use the half angle formula again and I am going to write the half angle formula over here.0491

But, I am going to use U this time.0496

1 + cos(2u)/2 because here we have cos2(2x), but if you think of that as being U, you can convert that into 1/4 the integral of 1 - cos2(u).0500

cos2(u) is 1 + cos(2u).0523

1 + cos, now 2u if u is 2x, u is 4x.0530

This is 1/4 × cos(4x) dX.0540

That simplifies a little bit to 1/4 × 1/2 - 1/2 sin(4x) dx.0558

I am going to pull those halves out and combine them with the 1/4, and we get 1/8.0573

Times the integral of 1 - cos(4x) dX, and that is 1/8.0579

Well the integral of 1 is just x.0591

The integral of cosine is sine.0596

But, because it is 4x, we have to multiply on a 1/4.0601

Finally, we get (1/8)x - 1/8 × 1/4 = (1/32)sin(4x) and at the end of these we always add a constant.0606

In that one, we had to cope with the fact that there were no odd powers.0626

Both were even and we used the half angle formula to resolve that.0632

Sometimes after you use the half angle formulas, you find yourself with another even power as we did.0634

And so you have to use the half angle formula again to resolve that.0642

If you keep on using them, eventually you get down to single powers of cosine and that is something that you can integrate without too much trouble.0647

Let us look at another example.0661

This one is in terms of secants and tangents.0663

That is the other common pattern that you are going to see with trigonometric integrals.0665

That is, powers of secant(x), and tangent(x).0671

To do this, it is helpful to remember that the derivative of tan(x) is sec2(x).0674

The derivative of sec(x) is sec(x)tan(x).0685

What we see here is again, looking at these powers, I look at the secant power and I see that that is even.0699

The reason that that is significant is that I can separate that out into sec2(x) tan2(x) dx.0709

And, I remember that the derivative of tangent is sec2(x).0724

If I use the substitution u = tan(x) than dU is sec2(x) dX.0730

The point of this is that I can convert this integral into, I have got sec2(x) dX, so that is dU, tan2(x) is u2.0744

Tan2(x) + 1 is sec2(x).0764

Sec2(x) can be written as tan2(x) + 1 which I am going to convert directly into u2 + 1.0773

This integral converts into u3, sorry, u4 + u2 dU.0786

Again, that is a very easy integral.0800

That just integrates to u5/5 + u3/3.0803

Then we can substitute back, u is tan(x), so that is tan(x5/5) + tan(x3/3) + C.0808

Let us go back and look at what made that work.0831

What made that work again was the even power on sec(x).0833

If you have an even power on sec(x), then you can always use this substitution u = tan(x) dU = sec2(x).0841

That even power of sec2(x) guarantees you 2 secants to use as the dU,0850

and then the other secants you can convert into tangents using the pythagorean identity.0856

The other place you can use substitution like this is if the power on tangent is odd, then you can use the other substitution u equals sec(x).0861

Then your dU will be sec(x)tan(x) dX.0892

The point of that is that you will have an odd tangent.0900

Because you have an odd tangent, you will use this pythagorean identity to convert the tangents into secants two at a time.0907

Starting with an odd secant you will be left with exactly 1 tangent left and that tangent you can save that to be your dU.0918

We will see another example of that later on.0930