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For the second integral, we complete the sqaure, make a shift , and use the trigonometric
substitution . We have , and we use the identity .
Since we have an even power of sine and cosine, we try to use the double angle formula,
.
To return to our original variable , we need to construct the triangle deﬁned by We see that so that
12
.
WATERLOO SOS E XAM AID: MATH138 M IDTERM
Then
We have
where is a constant. E XAMPLE 7. Calculate
S OLUTION. By inspection or using the substitution , we get
Wait! The Fundamental Theorem of Calculus only gives this relationship between integrals and derivatives
if the integrand is continuous on the closed interval , which is not true here. This is a trick question:
we must use the deﬁnition of improper integrals to attempt this problem, which is coming up next. E XERCISE 1. Calculate ...
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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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