A mixture of strategies
Last week, we learnt about
the fundamental theorem of calculus (part 2)
and techniques such as
substitution rule
integration by parts
that help compute definite and indefinite integrals. An appropriate
combination of these strategies, in conjunction with algebraic and trig
identities, prove to be surprisingly effective in handling a range of
integrands, involving
p
p
a2 ± x 2 ,
x 2 − a2 .
sinm x cosn x, secm x tann x,
One goal this week: Review the old concepts in the context of these
special integrands.
Math 105 (Section 204)
Trigonometric integrals
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Example 1: sinm x cosn x, at least one of m, n odd
Evaluate
Z
π
2
sin4 x cos3 x dx
0
(Hint: Use
cos2 x
2
= 1 − sin x.)
A. 1/5
B. 3/7
C. 1/2
D. 2/35
Math 105 (Section 204)
Trigonometric integrals
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Antiderivatives of tan x and sec x
Applications of substitution rule
Z
tan x dx = ln | sec x| + C ,
1.
Z
2.
Math 105 (Section 204)
sec x dx = ln | sec x + tan x| + C .
Trigonometric integrals
2011W T2
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Antiderivatives of tan x and sec x
Applications of substitution rule
Z
tan x dx = ln | sec x| + C ,
1.
Z
2.
sec x dx = ln | sec x + tan x| + C .
tan x =
sin x
cos x
1. Write
Math 105 (Section 204)
and substitute
Trigonometric integrals
u = cos x.
2011W T2
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Antiderivatives of tan x and sec x
Applications of substitution rule
Z
tan x dx = ln | sec x| + C ,
1.
Z
2.
sec x dx = ln | sec x + tan x| + C .
tan x =
sin x
cos x
1. Write
2. Write
sec x =
and substitute
u = cos x.
sec x(sec x + tan x)
sec x tan x + sec2 x
=
,
(sec x + tan x)
(sec x + tan x)
and substitute u = sec x + tan x!
Math 105 (Section 204)
Trigonometric integrals
2011W T2
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Example 2: secm x tann x, m odd, n even
Evaluate
Z
π
4
sec3 x dx.
0
A. (ln |secx + tan x|)3 + C
√
√
B. 22 + 12 ln( 2 + 1)
√
3
C. ln( 2 + 1)
D.
π
4
Math 105 (Section 204)
Trigonometric integrals
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Example 3: secm x tann x; m, n odd
Find f ( π4 ) if
f 0 (x) = sec3 x tan x,
A.
√
2
3(
1
f (0) = .
3
2 + 1)
B. undefined
C. 1
D. 2/3
Math 105 (Section 204)
Trigonometric integrals
2011W T2
5/5