Trigonometric Integrals and Substitution

Lecture 26 18.01 Fall 2006 Lecture 26: Trigonometric Integrals and Substitution Trigonometric Integrals � How do you integrate an expression like sinnx cosmx dx? (n = 0, 1, 2... and m = 0, 1, 2, . . .) We already know that: � sin x dx = − cos x + c � cos x dx = sin x + c and Method A Suppose either n or m is odd. � Example 1. sin3x cos2x dx. Our strategy is to use sin2x + cos2x = 1 to rewrite our integral in the form: � � 3 2 sin x cos x dx = f (cosx) sinx dx Indeed, � sin3x cos2x dx = � sin2x cos2x sin x dx = � (1 − cos2x) cos2x 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 − u2 )u2 (−du) 1 1 1 1 (−u2 + u4 )du = − u3 + u5 + c = − cos3 u + cos5 x + c 5 5 3 3 Example 2. � cos3 x 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 cos3 x dx = (1 − u2 )du = u − + c = sin x − +c 3 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 2x 2 cos2 x = (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) 2 cos x dx = dx = + +c 2 2 4 We follow a similar process for integrating sin2 x. sin2 x = � 2 � sin x dx = 1 − cos(2x) 2 1 − cos(2x) x sin(2x) 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. sin2 x cos2 x dx. Applying Method B twice yields �� � � � � � � 1 − cos 2x 1 + cos 2x 1 1 2 − cos 2x dx dx = 2 2 4 4 � � � 1 1 1 1 − (1 + cos 4x) dx = x − = sin 4x + c 4 8 8 32 There is a shortcut for Example 3. Because sin 2x = 2 sin x cos x, � 2 2 � � sin x cos x dx = 1 sin 2x 2 �2 1 dx = 4 � 1 − cos 4x dx = same as above 2 The next family of trig integrals, which we’ll start today, but will not finish is: � secn x tanm x 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 = dx = − = − ln(u) + c cos x u � tan x dx = − ln(cos x) + c � sec x dx is later.) (We’ll figure out what Now, let’s see what happens when you have an even power of secant. (The case n even.) � � � sec4 x dx = f (tanx) sec2 x dx = (1 + tan2 x) sec2 x dx Make the following substitution: u = tan x and du = sec2 x dx � sec4 x dx = � (1 + u2 )du = u + u3 tan3 x + c = tan x + +c 3 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, � 3 tan x secx dx = � (u2 − 1)du = u3 sec3 x − sec x + c −u+c= 3 3 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 � � sec x dx = du = 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 a Figure 1: Graph of the circle x2 + y2 = a2 . 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? x 0 Figure 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, � a2 − t2 = a2 − a2 sin2 u = a2 cos2 u; a2 − t2 = a cos u Plugging this into the integral gives us x � � � � a2 − t2 dt = (a cos u) a cos u du = a2 0 u=sin−1 (x/a) cos2 u du u=0 Here’s how we calculated the new limits of integration: � x � t = 0 =⇒ a sin u = 0 =⇒ u = 0 t = x =⇒ a sin u = x =⇒ u = sin−1 (x/a) a2 − t2 dt = a2 0 sin−1 (x/a) � cos2 u du = a2 0 2 = −1 a sin 2 (x/a) � + 2 a 4 � � � u sin 2u + 2 4 � � −1 �sin (x/a) � 0 � 2 sin(sin−1 (x/a)) cos(sin−1 (x/a)) (Remember, sin 2u = 2 sin u cos u.) We’ll pick up from here next lecture (Lecture 28 since Lecture 27 is Exam 3). 5