For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

## Discussion

## Download Lecture Slides

## Table of Contents

## Transcription

### 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

### AP Calculus AB Online Prep Course

### Transcription: More Example Problems for Trigonometric Integrals

*Hello, welcome back to www.educator.com, 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 www.educator.com.*1039

*We will see you next time, bye.*1041

1 answer

Last reply by: Professor Hovasapian

Mon Feb 8, 2016 1:36 AM

Post by Gautham Padmakumar on January 25, 2016

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