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# Chapter A Denite Integral Whose Indenite Form Cannot be Done As you know by now evaluating integrals is harder than computing derivatives PDF document - DocSlides

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Chapter 7 A Denite Integral Whose Indenite Form Cannot be Done As you know by now, evaluating integrals is harder than computing derivatives. An- other complication is that there are functions ), written in terms of functions you know, whose integrals cannot be written in terms of functions you know. To complicate things further, even if dx is such a function, it may be possible to evaluate dx for some values of and . Of course there are trivial cases such as when the integral is zero. What about nontrivial examples? One of the most popular ones to show students is 1 dx . This is done using multiple integrals, so it can't be done in this course. Here is an example that can be done: = ln(sin dx: (1) You might like to try doing ln(sin dx . You'll nd that you cannot. The integral in (1) is improper because the function is innite at = 0. In this paragraph, we show that it converges. Since sin x= for 0 = 2, we have ln(sin ln(2 x= ). Thus ln(sin ln + ln(2 = Since ln xdx ln 1, you should be able to show that = ln(sin dx converges. We now evaluate the integral using sin = 2 sin( x= 2) cos( x= 2) and substitution. First = ln(sin dx = ln 2 sin( x= 2) cos( x= 2) dx = ln 2 dx = ln(sin( x= 2)) dx = ln(cos( x= 2) dx: Now let x= 2= in the last two integrals to obtain = ln(sin dx =( ln 2) 2+2 = ln(sin du +2 = ln(cos du: Now let = in the last integral to obtain = ln(cos du = = ln(sin dt = = ln(sin dt; since cos( ) = cos cos + sin sin . Putting all this together we have = ln(sin dx =( ln 2) 2+2 = ln(sin du +2 = = ln(sin du =( ln 2) 2+2 = ln(sin dx: Hence the value of the integral in (1) is ln 2) 2.

An other complication is that there are functions written in terms of functions you know whose integrals cannot be written in terms of functions you know To complicate things further even if dx is such a function it may be possible to evaluate dx f ID: 23301

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Chapter 7 A Denite Integral Whose Indenite Form Cannot be Done As you know by now, evaluating integrals is harder than computing derivatives. An- other complication is that there are functions ), written in terms of functions you know, whose integrals cannot be written in terms of functions you know. To complicate things further, even if dx is such a function, it may be possible to evaluate dx for some values of and . Of course there are trivial cases such as when the integral is zero. What about nontrivial examples? One of the most popular ones to show students is 1 dx . This is done using multiple integrals, so it can't be done in this course. Here is an example that can be done: = ln(sin dx: (1) You might like to try doing ln(sin dx . You'll nd that you cannot. The integral in (1) is improper because the function is innite at = 0. In this paragraph, we show that it converges. Since sin x= for 0 = 2, we have ln(sin ln(2 x= ). Thus ln(sin ln + ln(2 = Since ln xdx ln 1, you should be able to show that = ln(sin dx converges. We now evaluate the integral using sin = 2 sin( x= 2) cos( x= 2) and substitution. First = ln(sin dx = ln 2 sin( x= 2) cos( x= 2) dx = ln 2 dx = ln(sin( x= 2)) dx = ln(cos( x= 2) dx: Now let x= 2= in the last two integrals to obtain = ln(sin dx =( ln 2) 2+2 = ln(sin du +2 = ln(cos du: Now let = in the last integral to obtain = ln(cos du = = ln(sin dt = = ln(sin dt; since cos( ) = cos cos + sin sin . Putting all this together we have = ln(sin dx =( ln 2) 2+2 = ln(sin du +2 = = ln(sin du =( ln 2) 2+2 = ln(sin dx: Hence the value of the integral in (1) is ln 2) 2.

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