Lecture_2 - CALCULUS II Spring 2009 LECTURE 2 This lecture...

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CALCULUS II Spring 2009 LECTURE 2 This lecture provides: more examples of integration by parts more examples of integration of trigonometric functions an introduction to integration by trig substitution a change of variables formula a remark that u-substitution and trig substitution are two different ways of looking at the change of variables formula. 1
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Recall We recall from Lecture 1: 10 examples of integration of trigonometric functions 4 examples of integration by parts End of Recall 2
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Mathematics Interlude I More Integration of Trigonometric Functions The next example rears its head in a substantial number of calculus model- ing problems. EXAMPLE. Evaluate the integral Z π 4 0 sec 3 ( x ) dx. Solution. Using the integration by parts formula Z b a udv = uv | b a - Z b a v du and the identity tan 2 θ = sec 2 ( θ ) - 1 , one writes Z π 4 0 sec 3 ( x ) dx = Z π 4 0 sec ( x ) sec 2 ( x ) dx = Z π 4 0 sec ( x ) d dx tan ( x ) dx = sec ( x ) tan( x ) | π 4 0 - Z π 4 0 tan ( x ) sec( x ) tan( x ) dx = ( 2 - 0) - Z π 4 0 sec ( x ) tan 2 ( x ) dx = 2 - Z π 4 0 sec ( x )(sec 2 ( x ) - 1) dx = 2 - Z π 4 0 sec 3 ( x ) dx + Z π 4 0 sec ( x ) dx. 3
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Since the integral R π 4 0 sec 3 ( x ) dx appears on both sides of the equation, one continues with the computation by adding R π 4 0 sec 3 ( x ) dx to both sides of the equation. One obtains: 2 Z π 4 0 sec 3 ( x ) dx = 2 + ln (sec ( x ) + tan ( x )) π 4 0 = 2 + [ln ( 2 + 1) - ln (1 + 0)] = 2 + ln ( 2 + 1) . Thus R π 4 0 sec 3 xdx = 2 2 + 1 2 ln ( 2 + 1). The next example is typical of integrations that need to be made in the study of Fourier Series. EXAMPLE. Evaluate the integral Z 1 0 sin (2 πx ) cos(3 πx ) dx. Solution. One writes Z 1 0 sin (2 πx ) cos(3 πx ) dx = 1 2 Z 1 0 [sin (5 πx ) + sin ( - πx )] dx = 1 2 Z 1 0 [sin (5 πx ) - sin ( πx )] dx = 1 2 - cos (5 πx ) 5 π ± ± ± ± 1 0 + cos ( πx ) π ± ± ± ± 1 0 ! = 1 2 ² 1 5 π + 1 5 π ³ + 1 2 ² - 1 π - 1 π ³ = 1 5 π - 1 π = - 4 5 π . sin ( x + y ) = sin ( x ) cos ( y ) + cos ( x ) sin ( y ) sin ( x - y ) = sin ( x ) cos ( y ) - cos ( x ) sin ( y ) add sin ( x ) cos ( y ) = 1 2 (sin ( x + y ) + sin ( x - y )) 4
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Thus R 1 0 sin (2 πx ) cos(3 πx ) dx = - 4 5 π . — END OF MATHEMATICS INTERLUDE — Mathematics Interlude II Change of Variables The change of variables formula plays a dominant role in the computation of integrals. We recall, once more, an exact statement. Theorem . Change-of-Variables . Let x ( t ) denote a real-valued differen- tiable function defined on an interval [ c,d ] and let a = x ( c ) and b = x ( d ). Let f denote a real-valued function defined and continuous on an interval I such that x ([ c,d ]) I . If x 0 ( t ) is continuous on [ c,d ] then Z b a f ( x ) dx = Z d c f ( x ( t )) x 0 ( t ) dt. If x ( t ) is chosen such that a = x ( d ) and b = x ( c ) then the change of variable formula becomes Z b a f ( x ) dx = Z c d f ( x ( t )) x 0 ( t ) dt. In stating the change-variables-theorem, we have used the symbol x to represent two different mathematical constructs. In the expression Z b a f ( x ) dx the symbol x represents an arbitrary point in the domain of the func- tion f . (The expression dx indicates that the integration takes place
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Lecture_2 - CALCULUS II Spring 2009 LECTURE 2 This lecture...

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