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E XAMPLE 5. Calculate
, for . S OLUTION. The strategy is not immediately obvious. There is no clear simpliﬁcation to be made, no trigonometric functions, no radicals, and no rational functions. We might be able to use integration by parts, using
the “1” trick. (Notice that the integrand is continuous for ). We have Then
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WATERLOO SOS E XAM AID: MATH138 M IDTERM E XAMPLE 6. Calculate
S OLUTION. We can attempt to simplify the integrand, using the double angle formula for sine, to get
There is no obvious antiderivative, nor can we immediately use a substitution like or
since we do not have an extra term to account for the change of variables. However, notice that we can
get a term appearing in the denominator if we divide through by , yielding a term in the
numerator. We then save a copy of , and express the remainder in terms of :
From here it is clear that a good substitution would be , with . Then
We can use a partial fraction decomposition now, say
which, after multiplying by , yields
Comparing coefﬁcients gives
Solving,
Then,
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WATERLOO SOS E XAM AID: MATH138 M IDTERM
For the ﬁrst integral, we complete the square and make a shift :
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This document was uploaded on 03/04/2014 for the course MATH 138 at Waterloo.
 Winter '07
 Anoymous
 Math, Calculus

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