Analytic Geometry and Calculus II - Module 1: Integration Summary

Calculusmodule21
1. Regular
2. Partial fraction
3. Trig integrals
4. Sub I
5. Trig substitution
6. By parts
7. Sub II
8. Improper integrals

, Indefinite integrals, substitution method
based on simple


Differentiation Formula Regular Integration derivative
Substitution (anti-chain)

(Sin x)
Sxnax = + cn =
(H) COS
S2x (x2 + 1)100dx
=
= 0 X .




nyn /secxdx
-
1
(xn)) = ((05X) = -
Sinx
Ex .
= tanx + c
1 le+ u = X
=
+1


lex) = eY (tanX)1 =
Sec2X don't forget ! * 2x(2x 4100ax
= .




(bY) = (Inb) b Y (se(X)1 = SecX . tan x (xax = (n(x) + c dx = 12x .


(Inx)) = ((0 + x)) = -
CSC2X (x2 + 1)10 + C
I


(109yY)1 =
(In b) x ((s(X) = -
(s(X(0 + X
I
J (x + 7)10dx = (x + 7)"#no chain rule Ssecx dx
(cos - x)
carccosx)"
-
=



1 -
x2
· Jeex .
Stand
+anx
Jse(2x
+ secX

((2x
I
(tan *) axz 1)1dx H ((2x 1)"
.



=
+ = + dx
+ ar secx + tanx
Carctanx)'
let u = SecX + tan x




Integration by parts


Su . du = u .
v-fv du . 1 Reduce Power Ex . 2 Eliminate In x Ex .




It cctdt, (x1nxdx
(t .
g)) = + g .
+ + g .
1 u = t dv = Csc2t 1u = 1nxdv = xdx

(4 .
V. = uv + u .
V du = 1 v = -
CO + t du = v =
Ex2
(u . ) = du . v + u . du -
+ cott -f -
co + + [xinx-(2x
((u .
v =
(v . du + Su . d -
- cott -

J- SOS Ex-(nx- [x
+
+ C


U = Sint (subI) 3 Recursive Ex .




du = costdX
Jet sinx dy

1 .
Reduce Power 1- cos =dx 1 u= eXdV = sinx



(xexdx(x2sinxdx -In It1 du du = eX v = -
COSX

u du u du
-
t cott -
( In(t)
-
+ C JeY sinx = -excosx +
Je Y
cosX

*
.
2 Eliminate In X ,
tan- X -
+co + t + In(t) + C U = e dv = COSX



(xinxax(x2 +an xax du = e
*
v = sinx

du
du
JeY sinx -excosx exsinx-Jexsinx
= +



.
3 Recursive 2)e" sinx excosx exsinx
= -
+




JexsinxdxJexcosxax(sec'xdx Je sinx -Le =
+ exsinx
*
cosx

u dr u du




More on Sub I

(x -
,
ax


le + u = x2 -
1 du = 2xdx
,




If u du


In (u1 + C


[in(x -
11 + c