The Free Intuitive Calculus Course Integrals

Transcription

1 Intuitive-Calculus.com Presents The Free Intuitive Calculus Course Integrals Day 19: Trigonometric Integrals By Pablo Antuna 013 All Rights Reserved. The Intuitive Calculus Course - By Pablo Antuna

2 Contents 1 Welcome Types of Trigonometric Integrals 3 3 First Case: Odd Powers of sinx or cosx 5 4 Second Case: Even Powers of sinx or cosx 7 5 Third Case: Integrals Involving tanx or secx 9 6 Fourth Case: Tricky Trigonometric Integrals 11 7 Exercises 13 8 Still To Come 14 1

3 1 Welcome Welcome to Day 19 of the Intuitive Online Calculus Course! The main purpose of this course is to give you the basic tools to succeed in calculus, whether you re in high school, college or self-studying calculus! Today we are going to learn how to solve trigonometric integrals.

4 Types of Trigonometric Integrals Our goal is to learn how to solve most trigonometric integrals. A goal of knowing how to solve all trigonometric integrals is not realistic, because there are some seemingly simple integrals that do no have simple functions as primitives. For example: sinx dx We are going to learn how to solve any integral that is solvable by hand. And to do that, it will be useful to classify trigonometric integrals into four cases. First Case: Integrals With Odd Powers of sinx or cosx For example: sin 3 xdx cos 4 x sin 3 x dx Second Case: Integrals With Only Even Powers of sinx or cosx For example: cos x dx Third case: Integrals Involving tanx or secx For example: tan 3 xdx 3

5 Fourth: Just Tricky Trigonometric Integrals For example: sinx cos 4 x + sin 4 x dx This last four case is for integrals that do not fit into any of the other cases. Now, let s solve an example of each of these. 4

6 3 First Case: Odd Powers of sinx or cosx Example 1 Let s solve the integral: sin 3 xdx Solution. We can write this integral as: sin 3 x dx = sin x sinx dx And use the identity: sin x = 1 cos x sin x sinx dx = (1 cos x ) sinx dx And we can break this up into two integrals: sinx dx cos x sinx dx The first integral can be solved directly, and the second one by substitution: u = cosx, sinx dx du dx = sinx cos x sinx dx = = cosx u. ( du ) dx.dx cos x sinx 5

7 That is: cosx + u. du dx. dx = cosx + u du cosx + u3 3 + C = cosx + cos3 x + C 3 sin 3 x dx = cosx + cos3 x 3 + C 6

8 4 Second Case: Even Powers of sinx or cosx Example Let s solve the integral: cos x dx Solution. When we have even powers of sinx or cosx, the trick is to use the following trig identity: cosx = cos(x + x) = cos x sin x That is: cosx = cos x 1 = 1 sin x cos x = 1 + cosx sin x = 1 cosx Using the first of these identities in our integral: 1 + cosx cos x dx = dx And we can now solve this integral by substitution: 1 + cosx u = x, du dx = 1 + cosu dx =. ( ) 1.du dx.dx 1 7

9 And we get: 1 + cosu. du 1 + cosu 4 dx. dx = du 4 And this integral is solved easily: 1 + cosu du = 1 du + 1 cos udu = u 4 sinu 4 + C And now we replace back u = x: cos x dx = x sinx 4 + C 8

10 5 Third Case: Integrals Involving tanx or secx Example 3 Let s solve the integral: tan 3 x dx Solution. We want to use the identity: And the fact that: 1 + tan x = sec x d dx (tanx) = sec x So, we write this integral as: tan 3 x dx = tan x tanx dx And replace tan x = sec x 1: (sec tan x tanx dx = 1 ) tanx dx = = tan x tanx sec x dx tanx dx sinx tanx sec x dx cosx dx 9

11 And we can solve both of these integrals by substitution: = u. u = tanx, v = cosx, du dx = sec x dv dx = sinx sinx tanx sec x dx cosx dx du dx.dx 1 v. ( dv ) dx.dx tanx sec x 1 cosx u. du 1 dx. dx + v.dv dx. dx dv udu + v = u + lnv + C And we substitute back u = tanx and v = cosx: tan 3 x dx = tan x + lncosx + C sinx 10

12 6 Fourth Case: Tricky Trigonometric Integrals Example 3 Let s solve the integral: sinx cos 4 x + sin 4 x dx Solution. In this case we can only guess what trigonometric identity will do the trick. So, first of all, we recognize that the denominator looks somewhat like a perfect square. For example, we have that: ( cos x + sin x ) = cos 4 x + sin x cos x + sin 4 x = 1 So: cos 4 x + sin 4 x = 1 sin x cos x = 1 sin x Here we used the identities: And: cos x + sin x = 1 sinx = sinx cosx sin x Replacing this in our integral: sinx cos 4 x + sin 4 x dx = sinx 1 sin x = sin x cos x dx = sinx sin x dx 11

13 Now, this looks a little more familiar. We can try the substitution: u = x, sinx sinx dx = du dx = sinu sin u. ( ) 1.du dx.dx 1 So, we must solve the integral: sinu sinu.1.du dx. dx = sinu sin u.du Now, what comes to mind, is using sin u = 1 cos u: sinu sin u.du = sinu 1 + cos u du = sinu 1 + cos u du And now we try the substitution: v = cosu, sinu 1 + cos u du = = dv du = sinu ( v. dv 1 + v dv du And this integral is solved directly, since (arctanv) = 1 dv 1 + v = arctanv + C And now, we substitute back v = cos u = cos x: ). du 1+v : sinx cos 4 x + sin 4 dx = arctancosx + C x 1

14 7 Exercises Solve the integrals: 1. cos 3 xdx. sin xdx 3. sin 4 xdx 4. cos 4 x sin 3 x dx dx 5. (this one is tricky!) (1 + cosx) 13

15 8 Still To Come Day 0: Introduction to Definite Integrals Day 1: The Fundamental Theorem of Calculus 14