2-4 - Last Time Two Techniques of Integration Last time we...

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Last Time: Two Techniques of Integration Last time we talked about two tools for integrating difficult functions: I Integration by Substitution (Change of Variable) Z x = b x = a f ( u ( x )) dx = Z u = u ( b ) u = u ( a ) f ( u ) dx du du = Z u = u ( b ) u = u ( a ) f ( u ) x 0 ( u ) du = Z u = u ( b ) u = u ( a ) f ( u ) u 0 ( x ) du I Integration by Parts Z udv = uv - Z vdu Z b a udv = uv b a - Z b a vdu
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Remember the following “tips” we talked about: I If you see a compound function, think about trying a substitution. Eg., Z 2 1 e 3 x dx = Z 6 3 e u 1 3 du = 1 3 Z 6 3 e u du I If you’re integrating the product of a polynomial times a function that anti-differentiates nicely (like e x or sin( x ) or cos( x )), try integration by parts. Eg., Z xe x dx = xe x - Z e x dx = xe x - e x + C I If you have a polynomial times ln( x ), ignore the last tip— differentiate the ln( x ) and anti-differentiate the polynomial. Eg., Z x 2 ln( x ) = x 3 3 ln( x ) - Z x 3 3 1 x dx = 1 3 x 3 ln( x ) - 1 3 Z x 2 dx
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Polynomial Substitutions— R 1 0 x 4 - x 2 dx First, notice the composite function in the denominator, 4 - x 2 . This immediately suggests trying the substitution u = 4 - x 2 , which gives du = - 2 xdx . Fortunately, we already have the requisite - 2 x in the numerator! (once we play with constants...) Z 1 0 x 4 - x 2 dx = - 1 2 Z 1 0 - 2 x 4 - x 2 dx = - 1 2 Z x =1 x =0 1 4 - x 2 · ( - 2 xdx ) = - 1 2 Z u =3 u =4 1 u du = - 1 2 Z u =3 u =4 u - 1 / 2 du = - 1 2 2 u 1 / 2 u =3 u =4 = - (3 1 / 2 - 4 1 / 2 ) = 2 - 3
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WARNING: As we already saw in the case of R 2 0 4 - x 2 dx , this “first instinct” substitution is not necessarily going to work! Z x =2 x =0 p 4 - x 2 dx = Z u =0 u =4 udx but there’s no good way to change from “ dx ” to “ du ” without an extra x factor. *Remember: Adding (or removing) constants (like - 2) is easy— but you can’t add or remove functions!
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Hidden Substitutions— R tan( x ) dx First, use a trigonometric identity to write Z tan( x ) dx = Z sin( x ) cos( x ) dx Now look at what we have: the expression looks more complicated, but the sin( x ) in the numerator is perfectly suited for the substitution u = cos( x ), since this means du = - sin ( x ) dx . Z tan ( x ) dx = Z sin( x ) cos( x ) dx = - Z 1 cos( x ) ( - sin( x ) dx ) = - Z 1 u du = - ln | u | + C = - ln | cos( x ) | + C = ln | sec( x ) | + C
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Sines and Cosines— R sin 3 ( x ) cos 2 ( x ) dx We know how to integrate functions like sin n ( x ) cos( x ), or sin( x ) cos m ( x ), by splitting off the solitary sin( x ) or cos( x ) to make a substitution. For example, Z sin 3 ( x ) cos( x ) dx = Z sin 3 ( x ) · (cos( x ) dx ) = Z u 3 du = 1 4 u 4 + C = 1 4 sin 4 ( x ) + C by making the substitution u = sin( x ) (so that du = cos( x ) dx ). But what if both sin( x ) and cos( x ) are raised to a power?
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