lec26 - Lecture 26 18.01 Fall 2006 Lecture 26:...

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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 x dx ? ( n = 0 , 1 , 2 ... and m = 0 , 1 , 2 , . . . ) We already know that: sin x dx = - cos x + c and cos x dx = sin x + c Method A Suppose either n or m is odd. Example 1. sin 3 x cos 2 x dx . Our strategy is to use sin 2 x + cos 2 x = 1 to rewrite our integral in the form: sin 3 x cos 2 x dx = f (cos x ) sin x dx Indeed, ± ± ± sin 3 x cos 2 x dx = sin 2 x cos 2 x sin x dx = (1 - cos 2 x ) cos 2 x sin x dx Next, use the substitution u = cos x and du = - sin x dx Then, ± ± (1 - cos 2 x ) cos 2 x sin x dx = (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 x dx = f (sin x ) cos x dx = (1 - sin 2 x ) cos x dx Again, use a substitution, namely u = sin x and du = cos x dx u 3 sin 3 x cos 3 x dx = (1 - u 2 ) du = u - + c = sin x - + c 3 3 1
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± ± ± ± ² ³ ± ± ± 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
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lec26 - Lecture 26 18.01 Fall 2006 Lecture 26:...

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