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2-9 - Last Week Two Techniques of Integration Last week we...

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Last Week: Two Techniques of Integration Last week 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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More Tips We saw some instances where it was not obvious which tool(s) to use. I Trigonometric Identities: Eg., I Z tan( x ) dx = Z sin( x ) cos( x ) dx I Z sin 2 ( x ) dx = Z 1 2 - cos(2 x ) 2 dx I Z sin 3 ( x ) dx = Z (1 - cos 2 ( x )) sin( x ) dx which lead to a natural substitution.
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More Tips I The “ · 1” Trick , eg. Z tan - 1 ( x ) dx = Z tan - 1 ( x ) · 1 dx = x tan - 1 ( x ) - Z x · 1 1 + x 2 dx by integration by parts, and the integral R x 1+ x 2 dx on the right can be solved with the substitution u = 1 + x 2 . I Cyclic Integration by Parts, eg. Z sin( x ) cos( x ) dx = sin 2 ( x ) - Z cos( x ) sin( x ) dx 2 Z sin( x ) cos( x ) dx = sin 2 ( x ) + C Z sin( x ) cos( x ) dx = 1 2 sin 2 ( x ) + C
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More Techniques Let’s consider three similar looking— but very different!— integrals: I Z 1 0 2 x 1 + x 2 dx = Z 2 1 du u = ln(2) Can be handled with the substitution u = 1 + x 2 , since du = 2 xdx is waiting for us in the numerator. I Z 1 0 2 1 + x 2 dx = 2 tan - 1 ( x ) 1 0 = π 2 Cannot be solved with the substitution u = 1 + x 2 , but there is a “trigonometric substitution” we can do. I Z 1 0 2 1 - x 2 dx = (diverges) Substitutions will not help you here! (At least, not directly.) This integral requires the method of “partial fractions”.
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Trigonometric Substitutions The name of the game is the following trigonometric identities: I 1 - sin 2 ( θ ) = cos 2 ( θ ) I 1 + tan 2 ( θ ) = cos 2 ( θ ) cos 2 ( θ ) + sin 2 ( θ ) cos 2 ( θ ) = sec 2 ( θ ) I sec 2 ( θ ) - 1 = tan 2 ( θ ) Notice that in each case, the right hand side of the identity is a perfect square.
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Trigonometric Substitution— R 1 1+ x 2 dx Remember the identity 1 + tan 2 ( θ ) = sec 2 ( θ ) Maybe we could use the substitution x = tan( θ ) to turn the denominator 1 + x 2 into sec 2 ( θ )? This will turn out to be useful because dx = sec 2 ( θ ) d θ . Watch: Z 1 1 + x 2 dx = Z 1 1 + tan 2 ( θ ) · sec 2 ( θ ) d θ = Z 1 sec 2 ( θ ) sec 2 ( θ ) d θ = Z d θ = θ + C = tan - 1 ( x ) + C
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Trigonometric Substitution— R 1 0 1 - x 2 dx We already met this integral, and we know it gives the area of one-quarter of the unit circle. This time, remember the identity 1 - sin 2 ( θ ) = cos 2 ( θ ) q 1 - sin 2 ( θ ) = | cos( θ ) | So perhaps it makes sense to use the change of variables x = sin( θ ) dx = cos( θ ) d θ This gives Z x =1 x =0 p 1 - x 2 dx = Z θ = π/ 2 θ =0 q 1 - sin 2 ( θ ) · (cos( θ ) d θ ) = Z θ = π/ 2 θ =0 | cos( θ ) | · cos( θ ) d θ = Z θ = π/ 2 θ =0 cos 2 (
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