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CHAPTER 7. TECHNIQUES OF INTEGRATION 7.2 33 Trigonometric Integrals Example 1. Find � sin(x) cos18 (x) dx. Solution. Let u = cos(x), du = − sin(x) dx and the integral becomes � − u18 du Then the anti-derivative is − Example 2. Find u19 cos19 =− 19 19 � sin7 (x) � sin2 (x) cos2 (x) dx � cos(x) dx. Solution. Let u = cos(x), du = sin(x) dx. We get rid of sin6 (x) by rewriting it as (1 − cos2 (x))3 . Then we have: � � � 7 sin (x) cos(x) dx = sin(x)(1 − cos2 (x))3 (cos(x))1/2 dx � = − (1 − u2 )3 u1/2 du � = − (1 − 3u2 + 3u4 − u6 )u1/2 du � = − u1/2 − 3u5/2 + 3u9/2 − u13/2 du � � 2 3/2 6 7/2 6 11/2 2 15/2 =− u − u + u − u 3 7 11 15 � � 2 6 6 2 3/2 7/2 11/2 15/2 =− (cos(x)) − (cos(x)) + (cos(x)) − (cos(x)) 3 7 11 15 Example 3. Find Solution. We apply the identities mentioned in Case 3 above to get � � 1 1 1 (1 − cos(2x)) (1 + cos(2x)) dx = 1 − cos2 (2x) dx 2 2 4 � 1 1 1 − (1 + cos(4x)) dx = 4 2 � 1 1 1 = − cos(4x) dx 4 2 2 � 1 = 1 − cos(4x) dx 8 � � 1 sin(4x) = x− 8 4 CHAPTER 7. TECHNIQUES OF INTEGRATION Example 4. Find � 34 tan3 (x) sec3 (x) dx. Solution. Note that the power of tangent is odd, so we use u = sec(x) du = sec(x) tan(x) dx If we apply this we can translate part of the integral already � � 2 2 tan (x) tan(x) sec(x) sec (x) ()dx = tan2 (x)u2 du � �� � � �� � ���� u2 du du Now to finish we need to get rid of the last functions of u. We do this with the identity � � 2 2 tan (x)u du = (sec2 (x) − 1)u2 du � = (u2 − 1)u2 du � = u4 − u2 du u5 u3 − 5 3 sec5 (x) sec3 (x) = − 5 3 = Example 5. Find � sec3 (x) dx. Solution. We have an even number of tangents (zero) and an odd number of secants. The integral is already written in terms of secants, so we apply integration by parts. To do this, we need to write sec3 (x) as a product, namely as sec(x) sec2 (x). � � �� � � � �� � � �� � sec(x) sec (x) dx = sec(x) tan(x) − sec(x) tan(x) tan(x) dx � = sec(x) tan(x) − sec(x) tan2 (x) dx � = sec(x) tan(x) − sec(x)(sec2 (x) − 1) dx � � 3 = sec(x) tan(x) − sec (x) + sec(x) dx � � 3 sec (x) dx = sec(x) tan(x) − sec3 (x) dx + ln | sec(x) + tan(x)| � 2 sec3 (x) dx = sec(x) tan(x) + ln | sec(x) + tan(x)| anti-derivative 2 derivative anti-derivative CHAPTER 7. TECHNIQUES OF INTEGRATION � sec3 (x) dx = Example 6. Find � 35 � 1� sec(x) tan(x) + ln | sec(x) + tan(x)| 2 tan2 (x) sec(x) dx. Solution. Since we have an even number of tangents, and an odd number of secants, we start by getting rid of all powers of tangent � � � 2 2 tan (x) sec(x) dx = (sec (x) − 1) sec(x) dx = sec3 (x) − sec(x) dx. � The integrals are now known: sec(x) dx was given above, and the previous � example solved sec3 (x) dx. We put this in, simplify, and we are done: � 1� sec(x) tan(x) + ln | sec(x) + tan(x)| − ln | sec(x) + tan(x)| 2 � 1� = sec(x) tan(x) − ln | sec(x) + tan(x)| 2