Lecture 4: Techniques for Integration
II. part II: Integral of the tan x, sec x
Recall
(tan x)0 = sec2 x
1 + tan2 x = sec2 x
(sec x)0 = sec x tan x
R
tan x dx = ln | sec x| + c = − ln | cos x| + c
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ex.
sec2 x dx
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sec2 x tan3 x dx
,
=
=
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ex.
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sec x tan x dx
,
=
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tan2 x dx,
sec3 x tan x dx
=
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tan3 x sec x dx,
=?
1
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sec x dx,
sec3 x dx
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tanm x secn xdx
(1) n is even: Isolate a sec2 x ,rewrite the rest
sec x in terms of tan x, if needed.
Then use u−sub: let u = tan x.
(2) m is odd, m > 1, n 6= 0: Isolate a tan x sec x ,
set the rest tan x in terms of sec x, if needed.
Then use u−sub: let u = sec x.
(3) m is even: use
tan2 x = sec2 x − 1,
to convert the tanm x in terms of sec2 x.
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NOTE: Similar rules apply to
2
cotm x cscn xdx
Evaluate:
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ex. 1.
tan6 x sec4 xdx
7
( tan7
x
+
tan9 x
9
+ c)
(EVEN power of sec x )
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ex. 2.
tan3 x sec xdx
(ODD power of tan x )
3
3
( sec3
x
− sec x + c)
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ex. 3.
tan2 xdx
(EVEN power of tan x )
(tan x − x + c)
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ex. 4.
sec xdx
(ln | sec x + tan x| + c)
4
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ex. 5.
sec3 xdx
( 12 (sec x tan x + ln | sec x + tan x|) + c)
NOTE: In this class, we will learn how to evaluate the integrals of the sec x, tan x and csc x, cot x
without using a table of integrals.
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NYTI Integration gymnastics:
1. Similarly, you can apply the same technique
here to evaluate the integrals of the cosecant
and cotangent functions.
(cot x)0 = − csc2 x,
1+cot2 x = csc2 x
(csc x)0 = − csc x cot x
R
cot x dx = ln | sin x| + c
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ex. Evaluate
csc2 x dx
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,
=
csc2 x cot3 x dx
=
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ex. Evaluate
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csc x cot x dx
=
,
csc3 x cot x dx
=
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ex. Evaluate
csc xdx
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ex. Evaluate
csc3 xdx
6
(ln | csc x − cot x| + c)
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2.
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3.
tan4 xdx
( tan3
tan3 xdx
Idea:
3
x
− tan x + x + c)
( 21 tan2 x + ln | cos x| + c)
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4.
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5.
tan5 xdx
tan4 x
4
cos5 x
√
dx
sin x
−
tan2 x
2
√
2 sin x −
− ln| cos x| + c
4 sin5/2 x
5
+
2 sin9/2 x
9
+c
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6.
x sec x tan x dx
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7.
x sec x − ln| sec x + tan x| + c
cos x + sin x
dx
sin 2x
7
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8.
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9.
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10.
1 − tan2 x
dx
sec2 x
x cos2 x dx
cos x cos5(sin x) dx
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tan4 x tan6 x
tan3 x
dx =
+
+c
11. Show
cos4 x
4
6
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12.
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13.
dx
cos x − 1
x sec2(x2) tan4(x2) dx
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