Math 128 : Calculus 2 for the Sciences
Winter 2016
Lecture 4: January 11, 2016
Lecturer: Jen Nelson
4.1
Notes By: Harsh Mistry
Half Angle Identities
• cos2 x = 12 (1 + cos 2x)
• sin2 x = 12 (1 − cos 2x)
Example 4.1
Z
0
π
3
Z
π
3
1
(1 + cos 2x)dx
2
0
sin 2x π3
1
) |0
= (x +
2
2
√
cos2 xdx =
3
1 π
( + 2 )
2 3√ 2
π
3
= +
6
8
=
Example 4.2
Z
Z
1
sin2 x cos2 xdx =
(1 − cos 2x)(1 + cos 2x)dx
4
Z
1
=
(1 − cos2 2x)
4
Z
1
1
=
(1 − (1 + cos4 x))dx
4
2
Z
1
1 1
=
( − cos 4x)
4
2 2
sin 4x
1
)+C
= (x −
8
4
4.2
Integrals Involving Tan and Sec
Z
tanm x secn xdx, m, n ∈ Z, m, n ≥ 0
4-1
Lecture 4: January 11, 2016
4-2
Known Derivatives
d
d
tan x = sec2 x
sec x = sec x tan x
dx
dx
d
d
tanm x = m tanm−1 x sec2 x
secm = n secn−1 x sec x tan x = n secn x tan x
dx
dx
• Try to manipulate expression, so we have tan x or sec2 x
• Use identity 1 + tan2 x = sec2 x
Example 4.3 Z
Z
9
5
sec x tan xdx =
Z
=
Z
=
Z
=
Z
=
sec9 x tan4 x tan xdx
sec9 x(sec2 x − 1)2 tan xdx
sec9 x(sec 3x − 1)2 tan xdx
u8 (u2 − 1)2 du
u = sec x =⇒ du = sec x tan x
u8 (u4 − 2u2 + 1)du
u1 3 2u1 1 u9
−
+
+C
13
11
8
1
1
sec 3x 2 sec 1x sec9 x
−
+
+C
=
13
11
9
=
Example 4.4 Z
Z
sec4 x tan2 xdx = sec 3x tan2 x sec2 xdx
Z
= (1 + tan2 x) tan 3x sec2 xdx
Z
= (1 + u2 )u2 du
u = tan x =⇒ du = sec2 x
u3
u5
+
+C
3
5
tan3 x tan5 x
=
+
+C
3
5
=
Example 4.5 Z
Z
3
sec xdx = sec x tan x −
Z
= sec x tan x −
Z
= sec x tan x −
Z
1
sec3 xdx = sec x tan x +
2
tan2 x sec xdx
(sec2 x − 1) sec xdx
Z
3
sec xdx + sec xdx
1
ln | sec x + tan x | +C
2
u = sec x =⇒ du = sec x tan x
dv = sec2 xdx =⇒ v = tan x
Lecture 4: January 11, 2016
4-3
Note : Integrals involving Cot and Csc share a similar strategy
4.3
Trig Substitution
Interested in integrals containing
√
√
√
a2 − b2 x2 , a2 + b2 x2 or b2 x2 − a2
Example 4.6 Z
1
√
dx(= arcsin x + C)
1 − x2
Another Way to Solve
( :
x = sin β
Inverse Subsitution
dx = cos βdβ
Z
Z
cos
cos β
p
√
=
2
cos
3x
1 − sin β
End of Lecture Notes
Notes By : Harsh Mistry