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INSTRUCTORS Raffi Hovasapian John Zhu
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For more information, please see full course syllabus of AP Calculus AB
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Lecture Comments (2)

1 answer

Last reply by: Professor Hovasapian
Mon Feb 8, 2016 1:36 AM

Post by Gautham Padmakumar on January 25 at 10:35:31 AM

Hi Raffi do you not have videos on Inverse Trig Integrals?

More Example Problems for Trigonometric Integrals

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
  • Example I: ∫sin²(x)cos⁷(x)dx 0:14
  • Example II: ∫x sin²(x) dx 3:56
  • Example III: ∫csc⁴ (x/5)dx 8:39
  • Example IV: ∫( (1-tan²x)/(sec²x) ) dx 11:17
  • Example V: ∫ 1 / (sinx-1) dx 13:19

Transcription: More Example Problems for Trigonometric Integrals

Hello, welcome back to, and welcome back to AP Calculus.0000

Today, I thought we would actually do some more example problems for trigonometry integrals because they do tend to be tricky.0004

I just wanted to throw some examples in there, let us get started.0010

The first integral that we are asked to evaluate is the integral sin² cos⁷ dx.0016

Remembering or referring back to our summary of things, regarding sin to some power × cos to some power,0024

we see that the cos power is odd.0036

We are going to actually pull out the cos.0045

We are going to write this as the integral of sin² x cos⁶ x cos x dx.0048

Now we are going to write this as the integral of sin² x × cos² x³ cos x dx.0062

We are going to express cos² x, in terms of sin x.0078

Using the Pythagorean identity, sin² x 1 - sin² x³ cos x dx.0083

Yes, now I’m going to go ahead and use a u substitution.0101

I’m going to let u = sin x.0103

I’m going to let du = cos x dx.0107

What I end up with is the integral of u² 1 - u²³ du.0114

There we go, now let us multiply this out.0127

When I multiply this out, I end up with the integral of u² × 1 - 3u² + 3u⁴ – u⁶ du 0133

=the integral of u² – 3u⁴ + 3u⁶ - u⁸ du.0152

=u³/ 3 – 3u⁵/ 5 + 3u⁷/ 7 – u⁹/ 9 + c.0180

And then, you plug them back in.0201

Now you end up with, final answer is, the integral = sin³ x/ 3 - 3 sin⁵ x/ 5 + 3 sin⁷ x/ 7 – sin⁹ x/ 9 + c.0204

Example 2, we want to evaluate x sin² x dx.0238

I'm going to rewrite this integral as the integral of x and using that identity with sin²,0245

I’m going to write this as ½ × 1 - cos 2x dx.0253

This is going to equal the integral of x/ 2 - x cos 2x/ 2 dx = ½ × the integral of x dx - ½ × the integral of x cos x.0265

I already took care of the 2 here, x cos x dx.0295

This one is going to be ½ × x²/ 2 - this one, I will go ahead and just write -½ the integral of x cos x dx.0309

For this one, for this integral, we are going to use integration by parts because we have a function × another function.0330

I’m just going to go ahead and use tabular integration.0337

Integration by parts, but I’m going to do the quick version of it.0339

Never mind, I will do the full integration by parts.0343

u is equal to x, dv is equal to cos x, du = 1, the integral cos x = sin x.0348

Therefore, this integral ends up being uv.0365

The integral of u dv = uv - the integral of v du.0372

uv is going to be x sin(x) - the integral of v du - the integral of sin x × 1 dx.0377

We have the ½, I put the ½ here and I will put the ½ here.0392

We have x²/ 4 - ½ x.0408

I think it was sin 2x, was not it?0430

Yes, -1/2 × ½ x sin 2x - ½ the integral of sin(2x) dx = x²/ 4 - ¼ x sin(2x).0433

I think the strange feeling that I have made a mistake regarding some coefficient to ½ somewhere.0462

I think I either had too many or too little.0468

But the coefficients are ultimately kind of irrelevant.0470

It is the process that we are concerned with.0475

The integral of sin, ½ comes out.0480

½ × ½ is ¼, another half comes out.0488

This is going to be the integral of sin is – cos.0492

It is going to be + 1/8 cos(2x) + c.0495

There you go, hopefully I have not messed anything up.0509

Again, I think there might be an extra ½ somewhere that does not actually belong there, but it is not a problem.0513

How does one deal with this one, csc⁴ x/ 5.0524

Let us write this as the integral of csc² x/ 5 × csc² x/ 5 dx.0530

We know that 1 + cot² x = csc² x.0548

We are going to write this as 1 + cot² x/ 5 × csc² x/ 5 dx.0557

We will let u equal to cot x/ 5, du = -csc² x/ 5 × 1/5.0577

Therefore, -5 du = csc² x/ 5.0595

We have the integral of 1 + u² × -5 du.0607

It equal -5 × the integral of 1 + u² du which = -5 × u + u³/ 3.0620

Our final answer is going to be -5u which is the cot(x)/ 5 - 5 × cot³ (x)/ 5/ 3 + c.0644

There you go, that is it, just manipulation.0669

1 - tan²/ sec².0679

Let us separate this out, I should go back to black.0683

This is going to be, this/this and this/this.0688

It is the integral of 1/ sec² x dx - the integral of tan²/ sec² x dx.0691

This is equal to the integral of cos² x dx - the integral of sin² x/ cos² x/ 1/ cos² x dx.0710

Cos² go away, I'm left with the integral of cos² x dx - the integral of sin² x dx.0731

This is nice, you can do this as separate using those identities for sin² and cos².0746

Or you can recognize that this is just cos² – sin².0753

Cos² x – sin² x which is actually equal to the integral of cos 2x.0759

Cos² – sin² is double angle identity, cos 2x.0769

When you integrate this, integral of cos 2x is going to be ½ sin 2x + c.0776

There you go, nice.0786

Something that looks really complicated ends up being very simple.0787

You will never quite know.0793

Something simple, ends up being very complicated.0794

Example number 5, evaluate 1/ sin x -1.0800

When you see something like this, it is probably best to try to simplify.0808

I’m going to multiply by the conjugate of the denominator, just to make it something a little bit more tractable to deal with.0811

Let us do 1/ sin x - 1 × sin x + 1 sin x + 1.0818

When I multiply all that out, I’m going to end up getting sin x + 1/,0830

I will just go ahead and do it, it is not a problem.0839

We might as well over here, I’m not going anywhere.0842

I’m going to get sin x + 1/ sin² x - 1 which = sin x + 1.0845

Sin² x - 1 is equal to -cos² x.0861

This is actually equal to the integral of sin x + 1/ -cos² x dx.0868

What is the best way to handle this?0888

I’m going to separate this out as the integral of sin x/ -cos² x dx + the integral,0905

that is fine, let us do it that way.0919

1/ -cos² x dx = the integral of sin x/ -cos x.0924

I’m going to separate the cos² into cos x.0940

Cos x dx, pull this – out, - the integral of, I’m going to write it as sec².0943

1/ cos² is sec².0950

This is equal to sin x/ cos x.0954

I’m going to write is as –tan x.0959

-the integral of tan x, 1/ cos x is sec x.0961

The integral of sec tan is sec x.0974

The integral of sec² is tan x + c.0983

There you go, that is it, nice and straightforward.0989

The moral of all this is, basically there is no algorithmic approach.0993

I know that we summarize some things, when we have even powers of cosines and even powers of sine,0997

odd powers of cosine, odd powers of sine, and for the other functions of tan and sec.1002

But truth be told, aside from that and aside from some of the identities, 1008

it really is just a matter of rolling the dice 1012

and hoping to God that something that you try is actually going to bear some fruit.1016

No algorithmic approach, but you are used to that by now.1020

This is calculus, this is sophisticated mathematics.1022

Not moderately sophisticated, this is sophisticated mathematics and you know it is going to require some experimentation.1025

You will go this way, you will go that way.1033

You hit a wall, fine, go back to the beginning and start again, not a problem.1034

Thank you so much for joining us here at

We will see you next time, bye.1041