{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}