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§8.2–Trigonometric Integrals Mark Woodard Furman U Fall 2010 Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 1 / 14 Outline 1 Powers of sine and cosine Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 2 / 14 Powers of sine and cosine Essential identities Here are the basic identities that we need: Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 3 / 14 Powers of sine and cosine Essential identities Here are the basic identities that we need: cos2 (x) + sin2 (x) = 1 Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 3 / 14 Powers of sine and cosine Essential identities Here are the basic identities that we need: cos2 (x) + sin2 (x) = 1 1 + cos(2x) cos2 (x) = 2 Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 3 / 14 Powers of sine and cosine Essential identities Here are the basic identities that we need: cos2 (x) + sin2 (x) = 1 1 + cos(2x) cos2 (x) = 2 1 − cos(2x) sin2 (x) = 2 Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 3 / 14 Powers of sine and cosine Essential identities Here are the basic identities that we need: cos2 (x) + sin2 (x) = 1 1 + cos(2x) cos2 (x) = 2 1 − cos(2x) sin2 (x) = 2 1 + tan2 (x) = sec2 (x) Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 3 / 14 Powers of sine and cosine Powers of sine and cosine, Case 1 Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 4 / 14 Powers of sine and cosine Powers of sine and cosine, Case 1 Z Consider an integral of the form Mark Woodard (Furman U) cosm (x) sinn (x)dx: m or n is odd. §8.2–Trigonometric Integrals Fall 2010 4 / 14 Powers of sine and cosine Powers of sine and cosine, Case 1 Z Consider an integral of the form cosm (x) sinn (x)dx: m or n is odd. We use the identity sin2 (x) + cos2 (x) = 1 and u-substitution. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 4 / 14 Powers of sine and cosine Powers of sine and cosine, Case 1 Z Consider an integral of the form cosm (x) sinn (x)dx: m or n is odd. We use the identity sin2 (x) + cos2 (x) = 1 and u-substitution. Problem Z Evaluate I = sin4 (x) cos7 (x)dx. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 4 / 14 Powers of sine and cosine Solution Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 5 / 14 Powers of sine and cosine Solution Write sin4 (x) cos7 (x) = sin4 (x) cos2 (x) 3 cos(x) = sin4 (x) 1 − sin2 (x) Mark Woodard (Furman U) §8.2–Trigonometric Integrals 3 cos(x). Fall 2010 5 / 14 Powers of sine and cosine Solution Write sin4 (x) cos7 (x) = sin4 (x) cos2 (x) 3 cos(x) 3 = sin4 (x) 1 − sin2 (x) cos(x). Z Let u = sin(x). Then du = cos(x)dx and I = u 4 (1 − u 2 )3 du. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 5 / 14 Powers of sine and cosine Solution Write sin4 (x) cos7 (x) = sin4 (x) cos2 (x) 3 cos(x) 3 = sin4 (x) 1 − sin2 (x) cos(x). Z Let u = sin(x). Then du = cos(x)dx and I = u 4 (1 − u 2 )3 du. Then Z I = u 4 − 3u 6 + 3u 8 − u 10 du u 5 3u 7 3u 9 u 11 − + − +C 5 7 9 11 sin5 (x) 3 sin7 (x) 3 sin9 (x) sin11 (x) = − + − +C 5 7 9 11 = Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 5 / 14 Powers of sine and cosine Powers of sine and cosine, Case 2 Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 6 / 14 Powers of sine and cosine Powers of sine and cosine, Case 2 Z Consider an integral of the form Mark Woodard (Furman U) cosm (x) sinn (x)dx: m and n even. §8.2–Trigonometric Integrals Fall 2010 6 / 14 Powers of sine and cosine Powers of sine and cosine, Case 2 Z Consider an integral of the form cosm (x) sinn (x)dx: m and n even. Use the the half-angle identities, repeatedly if necessary. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 6 / 14 Powers of sine and cosine Powers of sine and cosine, Case 2 Z Consider an integral of the form cosm (x) sinn (x)dx: m and n even. Use the the half-angle identities, repeatedly if necessary. Problem Z Solve the integral π/2 sin2 (x)dx. 0 Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 6 / 14 Powers of sine and cosine Solution Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 7 / 14 Powers of sine and cosine Solution Z We will solve the anti-derivative problem I = sin2 (x)dx first and evaluate at the end. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 7 / 14 Powers of sine and cosine Solution Z We will solve the anti-derivative problem I = sin2 (x)dx first and evaluate at the end. Since sin2 (x) = (1 − cos(2x))/2, we have Z 1 − cos(2x) x sin(2x) I = dx = − + C. 2 2 4 Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 7 / 14 Powers of sine and cosine Solution Z We will solve the anti-derivative problem I = sin2 (x)dx first and evaluate at the end. Since sin2 (x) = (1 − cos(2x))/2, we have Z 1 − cos(2x) x sin(2x) I = dx = − + C. 2 2 4 We can drop the “+C” and evaluate the resulting anti-derivative between the limits, obtaining Z π/2 sin2 (x)dx = 0 Mark Woodard (Furman U) §8.2–Trigonometric Integrals π . 4 Fall 2010 7 / 14 Powers of sine and cosine Problem Z Evaluate I = sin2 (x) cos2 (x)dx. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 8 / 14 Powers of sine and cosine Problem Z Evaluate I = sin2 (x) cos2 (x)dx. Solution Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 8 / 14 Powers of sine and cosine Problem Z Evaluate I = sin2 (x) cos2 (x)dx. Solution By two applications of the double-angle formulas we find 1 − cos2 (2x) 4 1 cos(4x) = − . 8 8 sin2 (x) cos2 (x) = Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 8 / 14 Powers of sine and cosine Problem Z Evaluate I = sin2 (x) cos2 (x)dx. Solution By two applications of the double-angle formulas we find 1 − cos2 (2x) 4 1 cos(4x) = − . 8 8 sin2 (x) cos2 (x) = Finally Z I = Mark Woodard (Furman U) 1 cos(4x) − 8 8 dx = §8.2–Trigonometric Integrals sin(4x) x − +C 8 32 Fall 2010 8 / 14 Powers of sine and cosine Powers of tangent and secant, Case 1 Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 9 / 14 Powers of sine and cosine Powers of tangent and secant, Case 1 Z Consider an integral of the form tanm (x) secn (x)dx where n, the power of the secant, is even. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 9 / 14 Powers of sine and cosine Powers of tangent and secant, Case 1 Z Consider an integral of the form tanm (x) secn (x)dx where n, the power of the secant, is even. In this case, keep a sec2 (x) and convert the remaining secants to tangents through 1 + tan2 (x) = sec2 (x). Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 9 / 14 Powers of sine and cosine Powers of tangent and secant, Case 1 Z Consider an integral of the form tanm (x) secn (x)dx where n, the power of the secant, is even. In this case, keep a sec2 (x) and convert the remaining secants to tangents through 1 + tan2 (x) = sec2 (x). Make a u-substitution: u = tan(x), du = sec2 (x)dx. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 9 / 14 Powers of sine and cosine Powers of tangent and secant, Case 1 Z Consider an integral of the form tanm (x) secn (x)dx where n, the power of the secant, is even. In this case, keep a sec2 (x) and convert the remaining secants to tangents through 1 + tan2 (x) = sec2 (x). Make a u-substitution: u = tan(x), du = sec2 (x)dx. Problem Z Solve I = π/4 tan4 (x) sec6 (x)dx. 0 Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 9 / 14 Powers of sine and cosine Solution Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 10 / 14 Powers of sine and cosine Solution Write 2 tan4 (x) sec6 (x) = tan4 (x) sec2 (x) sec2 (x) 2 = tan4 (x) 1 + tan2 (x) sec2 (x). Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 10 / 14 Powers of sine and cosine Solution Write 2 tan4 (x) sec6 (x) = tan4 (x) sec2 (x) sec2 (x) 2 = tan4 (x) 1 + tan2 (x) sec2 (x). We make the substitution u = tan(x) and du = sec2 (x)dx; thus, Z 1 I = 4 2 2 Z u (1 + u ) du = 0 Mark Woodard (Furman U) 1 (u 4 + 2u 6 + u 8 )du = 0 §8.2–Trigonometric Integrals 143 315 Fall 2010 10 / 14 Powers of sine and cosine Powers of tangent and secant, Case 2 Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 11 / 14 Powers of sine and cosine Powers of tangent and secant, Case 2 Z Consider an integral of the form tanm (x) secn (x)dx where m, the power of the tangent, is odd. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 11 / 14 Powers of sine and cosine Powers of tangent and secant, Case 2 Z Consider an integral of the form tanm (x) secn (x)dx where m, the power of the tangent, is odd. In this case, keep one tangent and convert the remaining tangents to secants through 1 + tan2 (x) = sec2 (x). Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 11 / 14 Powers of sine and cosine Powers of tangent and secant, Case 2 Z Consider an integral of the form tanm (x) secn (x)dx where m, the power of the tangent, is odd. In this case, keep one tangent and convert the remaining tangents to secants through 1 + tan2 (x) = sec2 (x). Make a u-substitution: u = sec(x), du = sec(x) tan(x)dx. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 11 / 14 Powers of sine and cosine Powers of tangent and secant, Case 2 Z Consider an integral of the form tanm (x) secn (x)dx where m, the power of the tangent, is odd. In this case, keep one tangent and convert the remaining tangents to secants through 1 + tan2 (x) = sec2 (x). Make a u-substitution: u = sec(x), du = sec(x) tan(x)dx. Problem Z Solve π/3 tan7 (x) sec5 (x)dx. 0 Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 11 / 14 Powers of sine and cosine Solution Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 12 / 14 Powers of sine and cosine Solution Write 3 tan7 (x) sec5 (x) = tan(x) tan2 (x) sec5 (x) = tan(x) sec2 (x) − 1)3 sec5 (x) Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 12 / 14 Powers of sine and cosine Solution Write 3 tan7 (x) sec5 (x) = tan(x) tan2 (x) sec5 (x) = tan(x) sec2 (x) − 1)3 sec5 (x) If u = sec(x), then du = tan(x) sec(x)dx and 2 Z (u 2 − 1)3 u 4 du I = 1 2 Z (u 10 − 3u 8 + 3u 6 − u 4 )du = 1 4779 = 40 Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 12 / 14 Powers of sine and cosine A note on powers of tangents and secants Our analysis is not exhaustive: Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 13 / 14 Powers of sine and cosine A note on powers of tangents and secants Our analysis is not exhaustive: We do not have any direct methods for integrating powers of tangent alone and powers of secant alone, for example, Z Z 8 tan (x)dx and sec4 (x)dx. These integrals can be solved, but we will not address these cases by direct methods. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 13 / 14 Powers of sine and cosine A note on powers of tangents and secants Our analysis is not exhaustive: We do not have any direct methods for integrating powers of tangent alone and powers of secant alone, for example, Z Z 8 tan (x)dx and sec4 (x)dx. These integrals can be solved, but we will not address these cases by direct methods. We do not have any methods for integrating an even power of tangent times an odd power of secant, for example, Z tan4 (x) sec5 (x)dx. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 13 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C Show that R sec(x)dx = ln | sec(x) + tan(x)| + C Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C R Show that sec(x)dx = ln | sec(x) + tan(x)| + C R Evaluate sec2 (x)dx Hint: Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C R Show that sec(x)dx = ln | sec(x) + tan(x)| + C R Evaluate sec2 (x)dx Hint: Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C R Show that sec(x)dx = ln | sec(x) + tan(x)| + C R Evaluate sec2 (x)dx Hint: this is easy! Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C R Show that sec(x)dx = ln | sec(x) + tan(x)| + C R Evaluate sec2 (x)dx Hint: this is easy! R Evaluate tan2 (x)dx Hint: Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C R Show that sec(x)dx = ln | sec(x) + tan(x)| + C R Evaluate sec2 (x)dx Hint: this is easy! R Evaluate tan2 (x)dx Hint: Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C R Show that sec(x)dx = ln | sec(x) + tan(x)| + C R Evaluate sec2 (x)dx Hint: this is easy! R Evaluate tan2 (x)dx Hint: use a nice trig identity. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C R Show that sec(x)dx = ln | sec(x) + tan(x)| + C R Evaluate sec2 (x)dx Hint: this is easy! R Evaluate tan2 (x)dx Hint: use a nice trig identity. R Evaluate tan3 (x)dx Hint: Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C R Show that sec(x)dx = ln | sec(x) + tan(x)| + C R Evaluate sec2 (x)dx Hint: this is easy! R Evaluate tan2 (x)dx Hint: use a nice trig identity. R Evaluate tan3 (x)dx Hint: Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C R Show that sec(x)dx = ln | sec(x) + tan(x)| + C R Evaluate sec2 (x)dx Hint: this is easy! R Evaluate tan2 (x)dx Hint: use a nice trig identity. R Evaluate tan3 (x)dx Hint: save out a tangent and then use a trig identity. Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C Show that R Evaluate R Evaluate R Evaluate identity. R Evaluate R sec(x)dx = ln | sec(x) + tan(x)| + C sec2 (x)dx Hint: this is easy! tan2 (x)dx Hint: use a nice trig identity. tan3 (x)dx Hint: save out a tangent and then use a trig sec3 (x)dx Hint: Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C Show that R Evaluate R Evaluate R Evaluate identity. R Evaluate R sec(x)dx = ln | sec(x) + tan(x)| + C sec2 (x)dx Hint: this is easy! tan2 (x)dx Hint: use a nice trig identity. tan3 (x)dx Hint: save out a tangent and then use a trig sec3 (x)dx Hint: Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14 Powers of sine and cosine Problem (Some ad hoc problems) Show that R tan(x)dx = ln | sec(x)| + C Show that R Evaluate R Evaluate R Evaluate identity. R Evaluate R sec(x)dx = ln | sec(x) + tan(x)| + C sec2 (x)dx Hint: this is easy! tan2 (x)dx Hint: use a nice trig identity. tan3 (x)dx Hint: save out a tangent and then use a trig sec3 (x)dx Hint: try parts with u = sec x. (see page 500.) Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14