MATH101 Calculus Workshop Exercise 5 Solutions
W5.1. (i) The indefinite integral is
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(ii) Likewise we integrate each summand and obtain
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W5.2. The definite integrals are
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where the integral (ii) is similar to example in lectures.
W5.3. (i) We have
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hence we have
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(ii) By the Fundamental Theorem of Calculus and (i), we have
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W5.4. (i) We have
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(ii) We use the addition formulæ and obtain
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W5.5 (i) Integrating twice by parts, we have
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(ii) Likewise, we have
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W5.6 (i)
Integrating twice by parts, we have
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(ii) Using a hyperbolic identity, we obtain
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W5.7 Let
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(i) In particular we have
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(ii) We integrate by parts to get, for ,
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since We observe that
for , and that hence
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and we rearrange this to obtain the reduction formula
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By this reduction formula and the results of (i), we have
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please try to write this nicely
W5.8. (i) Now
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so
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(ii) We integrate by parts and obtain
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so, when we integrate by parts again, we obtain
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hence we have the recurrence relation
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Starting from the result for in lectures, we deduce
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W5.9. (i) Observe that , hence
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Alternatively, let , so , and substitute to get
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(ii) Observe that , so we
let and then , so
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W5.10. (i) By partial fractions, we have
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(ii) The partial fractions here give
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W5.11. Let so that and
Then
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and so we obtain by substituting
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where we recall the notation of lectures. In particular,
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W5.12. (i) Let , so that and .
The limits of integration change as
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Hence
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(ii) In this integral we let , so that and as before. The limits of
integration change as
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Hence
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W5.13. (i) We make the
substitution , so and find the integral
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(ii) In the integral
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we substitute , so and
; hence
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(iii) Now we look for constants and such that
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and then
so and ; hence
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W5.14. The difference of squares identity gives
, so
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where the final integral is given by guesswork or by using
the substitution .