Lecture Notes 26-38

Lecture Notes 26-38 - MIT OpenCourseWare http:/ocw.mit.edu...

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MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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Lecture 26 18.01 Fall 2006 Lecture 26: Trigonometric Integrals and Substitution Trigonometric Integrals How do you integrate an expression like sin n x cos m xdx ? ( n = 0 , 1 , 2 ... and m = 0 , 1 , 2 ,... ) We already know that: sin xdx = cos x + c and cos xdx = sin x + c Method A Suppose either n or m is odd. Example 1. sin 3 x cos 2 xdx . Our strategy is to use sin 2 x + cos 2 x = 1 to rewrite our integral in the form: sin 3 x cos 2 xdx = f (cos x ) sin xdx Indeed, sin 3 x cos 2 xdx = sin 2 x cos 2 x sin xdx = (1 cos 2 x ) cos 2 x sin xdx Next, use the substitution u = cos x and du = sin xdx Then, (1 cos 2 x ) cos 2 x sin xdx = (1 u 2 ) u 2 ( du ) 1 1 1 1 = ( u 2 + u 4 ) du = 3 u 3 + 5 u 5 + c = 3 cos 3 u + 5 cos 5 x + c Example 2. cos 3 xdx = f (sin x ) cos xdx = (1 sin 2 x ) cos xdx Again, use a substitution, namely u = sin x and du = cos xdx u 3 sin 3 x cos 3 xdx = (1 u 2 ) du = u + c = sin x + c 3 3 1
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± ² Lecture 26 18.01 Fall 2006 Method B This method requires both m and n to be even. It requires double-angle formulae such as 1 + cos 2 x 2 cos x = 2 (Recall that cos 2 x = cos 2 x sin 2 x = cos 2 x (1 sin 2 x ) = 2 cos 2 x 1 ) Integrating gets us 1 + cos 2 x x sin(2 x ) cos 2 xdx = dx = + + c 2 2 4 We follow a similar process for integrating sin 2 x . 1 cos(2 x ) sin 2 x = 2 1 cos(2 x ) x sin(2 x ) sin 2 xdx = dx = + c 2 2 4 The full strategy for these types of problems is to keep applying Method B until you can apply Method A (when one of m or n is odd). Example 3. sin 2 x cos 2 xdx . Applying Method B twice yields ± ²± ² ± ² 1 cos 2 x 1 + cos 2 x 1 1 2 2 dx = 4 4 cos 2 2 x dx 1 1 1 1 = 4 8 (1 + cos 4 x ) dx = 8 x 32 sin 4 x + c There is a shortcut for Example 3. Because sin 2 x = 2 sin x cos x , ± ² 2 sin 2 x cos 2 xdx = 1 sin 2 x dx = 1 1 cos 4 x dx = same as above 2 4 2 The next family of trig integrals, which we’ll start today, but will not finish is: sec n x tan m xdx where n = 0 , 1 , 2 ,... and m = 0 , 1 , 2 ,... Remember that sec 2 x = 1 + tan 2 x which we double check by writing 1 sin 2 x cos 2 x + sin 2 x = 1 + = cos 2 x cos 2 x cos 3 x sec 2 xdx = tan x + c sec x tan xdx = sec x + c 2
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Lecture 26 18.01 Fall 2006 To calculate the integral of tan x , write sin x tan xdx = dx cos x Let u = cos x and du = sin xdx , then sin x du tan xdx = cos x dx = u = ln( u ) + c tan xdx = ln(cos x ) + c (We’ll figure out what sec xdx is later.) Now, let’s see what happens when you have an even power of secant. (The case n even.) sec 4 xdx = f (tan x ) sec 2 xdx = (1 + tan 2 x ) sec 2 xdx Make the following substitution: u = tan x and du = sec 2 xdx u 3 tan 3 x sec 4 xdx = (1 + u 2 ) du = u + + c = tan x + + c 3 3
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This note was uploaded on 10/10/2011 for the course MATH 31 taught by Professor Blake during the Spring '11 term at MIT.

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Lecture Notes 26-38 - MIT OpenCourseWare http:/ocw.mit.edu...

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