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MIT 18 01 - Trigonometric Integrals and Substitution

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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.� � � � � � � � � Lecture 26 18.01 Fall 2006 Lecture 26: Trigonometric Integrals and Substitution Trigonometric Integrals How do you integrate an expression like sinn x cosmx 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. sin3 x cos2x dx. Our strategy is to use sin2 x + cos2x = 1 to rewrite our integral in the form: sin3 x cos2x dx = f(cosx) sinx dx Indeed, � � � sin3 x cos2x dx = sin2 x cos2 x sin x dx = (1 − cos2 x) cos2 x sin x dx Next, use the substitution u = cos x and du = − sin x dx Then, � � (1 − cos2 x) cos2 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 cos3 u + 5 cos5 x + c Example 2. � � � cos3x dx = f(sin x) cos x dx = (1 − sin2 x) cos x dx Again, use a substitution, namely u = sin x and du = cos x dx u3 sin3 x cos3x dx = (1 − u 2)du = u − + c = sin x − + c3 3 1� � � � � � � � � 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 2x2cos x = 2 (Recall that cos 2x = cos2 x − sin2 x = cos2 x − (1 − sin2 x) = 2 cos2 x − 1) Integrating gets us � � 1 + cos 2x x sin(2x)cos2 x dx = dx = + + c2 2 4 We follow a similar process for integrating sin2 x. 1 − cos(2x)sin2 x = 2 1 − cos(2x) x sin(2x)sin2x dx = dx = + c2 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. sin2 x cos2x dx. Applying Method B twice yields �� �� ��� �1 − cos 2x 1 + cos 2x 1 1 2 2 dx =4 − 4cos22x dx 1 1 1 1 =4 − 8(1 + cos 4x) dx =8 x − 32 sin 4x + c There is a shortcut for Example 3. Because sin 2x = 2 sin x cos x, � �� �2 � sin2 x cos2x dx = 1 sin 2x dx =1 1 − cos 4xdx = same as above 2 4 2 The next family of trig integrals, which we’ll start today, but will not finish is: secn x tanmx dx where n = 0, 1, 2, . . . and m = 0, 1, 2, . . . Remember that sec2 x = 1 + tan2 x which we double check by writing 1 sin2 x cos2 x + sin2 x= 1 + = cos2 x cos2 x cos3 x sec2 x dx = tan x + c sec x tan x dx = sec x + c 2� � � � � � � � � � � � � � � � Lecture 26 18.01 Fall 2006 To calculate the integral of tan x, write sin x tan x dx = dx cos x Let u = cos x and du = − sin x dx, then sin x du tan x dx = cos x dx = − u = − ln(u) + c tan x dx = − ln(cos x) + c (We’ll figure out what sec x dx is later.) Now, let’s see what happens when you have an even power of secant. (The case n even.) sec4x dx = f(tanx) sec2x dx = (1 + tan2 x) sec2x dx Make the following substitution: u = tan x and du = sec2x dx u3 tan3 x sec4x dx = (1 + u 2)du = u + + c = tan x + + c3 3 What happens when you have a odd power of tan? (The case m odd.) tan3 x sec x dx = f(sec x) d(sec x) = (sec2 x − 1) sec x tan x dx (Remember that sec2 x − 1 = tan2 x.) Use substitution: u = sec x and du = sec x tan x dx Then, � � u3 sec3 x tan3 x secx dx = (u 2 − 1)du =3 − u + c =3 − sec x + c We carry out one final case: n = 1, m = 0 sec x dx = ln (tan x + sec x) + c 3� � � � � � Lecture 26 18.01 Fall 2006 We get the answer by “advanced guessing,” i.e., “knowing the answer ahead of time.” sec x + tan x sec2 x + sec x tan x sec x dx = sec x dx = dx sec x + tan x tan x + sec x Make the following substitutions: u = tan x + sec x and du = (sec2 x + sec x tan x) dx This gives � � du sec x dx = = ln(u) + c = ln(tan x + sec x) + c u Cases like n = 3, m = 0 or more generally n odd and m even are more complicated and will be discussed later. Trigonometric Substitution Knowing how to evaluate all of these trigonometric integrals turns out to be useful for evaluating integrals involving square roots. Example 4. y = a2 − x2 a2 2Figure 1: Graph of the circle x2 + y = a . We already know that the area of the top half of the disk is a � πa2 a2 − x2 dx =2−a 4� � Lecture 26 18.01 Fall 2006 What if we want to find this area? 0xFigure 2: Area to be evaluated is shaded. To do so, you need to evaluate this integral: � t=x � a2 − t2 dt t=0 Let t = a sin u and dt = a cos u du. (Remember to change the limits of integration when you do a change of variables.) Then, � a 2 − t2 = a 2 − a 2 sin2 u = a 2 cos2 u; a2 − t2 = a cos u Plugging this into the integral gives us � x �� � u=sin−1(x/a) a2 − t2 dt = (a cos u) a cos u du = a 2 cos2 u du 0 u=0 Here’s how we calculated the new limits of integration: t = 0 = a sin u = 0 = u = 0 ⇒ ⇒ t = x = a sin …


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MIT 18 01 - Trigonometric Integrals and Substitution

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