MATH 12002 - CALCULUS I 5.6: More Integral Examples Professor - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I § 5.6: More Integral Examples Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 7

Integration Formulas � 1 � u n du = n +1 u n +1 + C 1 u du = ln | u | + C � � sin u du = − cos u + C cos u du = sin u + C � � 1 e u du = e u + C 1 − u 2 du = sin − 1 u + C √ � 1 � 1 1 + u 2 du = tan − 1 u + C du = sec − 1 u + C √ u 2 − 1 u D.L. White (Kent State University) 2 / 7

Examples � 1 � x 1 Evaluate 4 + 25 x 2 dx and 4 + 25 x 2 dx . We evaluated the first integral in the previous lecture: 4 + 25 x 2 dx = 1 1 1 2 x ) 2 dx = 1 � 5 � � � 10 tan − 1 2 x + C . 1 + ( 5 4 For the second, let u = 4 + 25 x 2 , so du = 50 x dx and 1 50 du = x dx : � 1 � 4 + 25 x 2 dx = 1 x u du = 1 50 ln | u | + C = 1 50 ln(4 + 25 x 2 ) + C . 50 D.L. White (Kent State University) 3 / 7

Examples e x e x � � 2 Evaluate √ 1 − e x dx and √ 1 − e 2 x dx . For the first integral, let u = 1 − e x , so du = − e x dx and ( − 1) du = e x dx : e x � � 1 � u − 1 / 2 du √ 1 − e x dx = − √ u du = − √ − 2 u 1 / 2 + C = − 2 1 − e x + C . = For the second integral, observe that e 2 x = ( e x ) 2 , and let u = e x , so du = e x dx : e x e x � � � 1 √ √ 1 − e 2 x dx = 1 − ( e x ) 2 dx = 1 − u 2 du � sin − 1 u + C = sin − 1 ( e x ) + C . = D.L. White (Kent State University) 4 / 7

Examples � sin x cos x � sin x cos 3 x 3 Evaluate 1 + cos 4 x dx and 1 + cos 4 x dx . To simplify these integrals, first let u = cos x so du = − sin x dx and ( − 1) du = sin x dx . We then have � sin x cos x � u 1 + cos 4 x dx = − 1 + u 4 du and � sin x cos 3 x u 3 � 1 + cos 4 x dx = − 1 + u 4 du . Each integral will now require a second substitution. [Continued → ] D.L. White (Kent State University) 5 / 7

Examples [Example 3, continued] For the integral � sin x cos x � u 1 + cos 4 x dx = − 1 + u 4 du , observe that u 4 = ( u 2 ) 2 and du u 2 = 2 u . d Now let t = u 2 so that dt = 2 u du and 1 2 dt = u du : � sin x cos x � u � u 1 + cos 4 x dx = − 1 + u 4 du = − 1 + ( u 2 ) 2 du − 1 1 � 2 tan − 1 t + C 1 + t 2 dt = − 1 = 2 − 1 2 tan − 1 ( u 2 ) + C = 2 tan − 1 (cos 2 x ) + C . − 1 = [Continued → ] D.L. White (Kent State University) 6 / 7

Examples [Example 3, continued] For the integral � sin x cos 3 x u 3 � 1 + cos 4 x dx = − 1 + u 4 du , let t = 1 + u 4 so that dt = 4 u 3 du and 1 4 dt = u 3 du : � sin x cos 3 x u 3 � 1 + cos 4 x dx = − 1 + u 4 du � 1 − 1 = t dt 4 − 1 = 4 ln | t | + C − 1 4 ln(1 + u 4 ) + C = 4 ln(1 + cos 4 x ) + C . − 1 = D.L. White (Kent State University) 7 / 7