Bonus Question: What substitution would you try for
Ksusha Gotham
Math 125
HW
/xdx
x2 1
+
reaa+b2Itanl
↓
1,
Video
↓
7 |
pw
43
·i
sidework
7.4
arctan(x)
dx
"(0)
COs
1
sin:
+
(8)
1
=
8
c
"(0)
1
+
-
csc
=
8d8
tan" (x)
=
or
I an"(0) (seC" OdE)
=
+
sec
=
dX
ran2(f) sx(t)
+
sec28 dE
(t)
X
variable
=
+
I
Pythagorean identity
reminder:
④
Le x tant
substitution
Trig
0.G
-
c
+
=
I
4siu8
=
dx 490S
=
=
(8)d8
I
Try using:
If denominator is:
az
·
-
x<
a.sinf
>X
=
az x2
+
.a,
y2
example:
>
X
a.tan(t)
=
austan
(x)
3.3: 126, 134 3.3 supplement: 1,3,6
Pythagorean identity
reminder:
3.3
4(1-sin" 8)
4-4sin" 8
126.
"(0)
COs
=
COS" A+sin? 8 1
sin:
+
(8)
1
=
ran2(f) sx(t)
1
+
=
=
c
COS28 1.sin"
"(0)
+
1
+
-
(t)
csc
=
=
substitute
COS'8
sin" as
1-
COS2 (8)
1
+
=
-
sin"
(8)
498SE
134.
I4
x
dx
az
-
b2
where
and b x
a 2
2sin(t)
->
=
=
Let
x
(8)
2 sin
=
(c'
-
dx
x2
dx
and
cos(8)dx
1
=
COS
↑
since
S: 'esinco (acoscolaxSc
=
2c0S(8)
-
sin
l
-
Xs
sin
Sa'itcosix 2cus(x)
=
(0s2x 1
+
Sacicosx)
2(0S(X)dx
X 2 sin8
=
Sc'pOX:(6x
=
0
=
+
arcsin(*) +
c
=
(*(2)
8:arcsin
3.3.51,3,6
"(+
x
(4)1
4/5sin(t)
2e x
=
+
at
·
2
dx
2/5C0s(t)dt
=
n(2/5c0s(t)) d
=
sin2t)
-
->
I
-
sin t
(2/5costdt
(OS
=
↓
1
1
+
CUSt
(2/5
(OS
(t)) d
cosst
4.2xzzsin"t)
I ify)
Iπ
(Esin(t)
sin"t
cos't
+
at
ost
.
=
SecosEd fset
=
1
2fse d+
=
=
I
3.
Let
sin27
1
4
-
x
25/sinit(
tan(t)
=
-
2/5(OSCZ
....
0(tan()
dx
16
(tan(t) 8)(tant
4
-
-cOS4t
Ysin't
2/5
cost
+
ElnDsect+an+1)
-
C
+
dx=
sec'tdx
8
x
-
-
I
=
1
=
t
nt
-
8
+
tan(t)
=
arctan (x
t
sec?tdx
-
-
1)
tant-stantant
3+ a
128
+64
arctay(x
=
=
I
I
8)
8)
+
+
↑
sc2tdt(d t+)
t
tan"t
-128
62) 65
+
I
tan?t
+
seid
1
+
->
sect
=
f/ secat
=
6.
/, y2dX
2e x 7+an(x)
+
ax
=
=
7s x
=
->
,
19
C
3
(7se2y)
49 7+an2x
sec2 tax" 1
+
=
+
([(z tan-x)
+
Sec2
1
=
-tan2
7sec2X
tan se,2 1
+
=
IAssecaxsec, dt E/secx
=
=
cot
be