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City College of New York MATH 21200 (Calculus 2 Notes) Page 1 of 3
th
Thomas’ Calculus Early Transcendentals, 14 ed Chapter 8 Section 1 to 5 author: Mr. Park
Section 8.1: Basic Integration Formulas
Formulas needed to be memorized: (*=absolutely needed)
p(1n+ ) 1
n * dp =ln p +C
pdp=+Cn≠−1
* ∫ n ∫ p
(1+ )
pp ap
* p
∫edp=+e C * ∫adp=lna+C
* *
∫sin pdp=−cos p+C ∫cos pdp=sin p+C
2 2
* *
∫sec pdp=+tan p C ∫csc pdp=−+cot p C
* sec pptan dp=+sec pC * cscppcot dp=−+cscpC
∫ ∫
* ∫csc p dp = −+ln csc p cot p +C
∫sec pdp=+ln sec p tan p+C *
=−ln csc cot +
p pC
* *
∫tan pdp=+ln sec p C ∫cot pdp=ln sin p+C
* *
∫sinh pdp=+cosh p C ∫cosh pdp=sinh p+C
1 p
11p −1 ⎛⎞
* −1 ⎛⎞ dp=sin +C
dp tan C
=+ ∫ ⎜⎟
∫ 22 ⎜⎟ 22 a
pa a a ⎝⎠
+ ap
⎝⎠ −
* 3 1 3 1
+=−+−+
sec pdp=+secptan p ln secp+tan p C csc pdp csc pcot p ln csc p cot p C
()()
∫ 2 ∫ 2
Section 8.2: Integration by Parts
The formula for integration by parts:
udv=−uv vdu
∫∫
st
Make sure to determine how you want to choose u and dv. Remember that your 1 choice might not be the
optimized method of solving by this technique.
There are problems that require you to apply integration by parts more than once.
Section 8.3: Trigonometric Integrals
Strategy for evaluating ∫sinm xcosn x dx
a) If the power of cosine is odd (n = 2k +1), then use cos2 x =1−sin2 x:
∫sinm xcos2k+1 x dx = ∫sinm x(cos2 x)k cosx dx = ∫sinm x(1−sin2 x)k cosx dx then let u = sinx
b) If the power of sine is odd (m = 2k +1), then use sin2 x =1−cos2 x:
∫sin2k+1 xcosn x dx = ∫(sin2 x)k sin xcosn x dx = ∫(1−cos2 x)k cosn xsin x dx then let u = cosx
c) If both powers of sine and cosine are even, then use half angle identities:
sin2 x = 1 (1−cos2x) cos2 x = 1 (1+ cos2x).
2 2
Sometimes this identity is helpful: sin xcosx = 1 sin2x
2
City College of New York MATH 21200 (Calculus 2 Notes) Page 2 of 3
th
Thomas’ Calculus Early Transcendentals, 14 ed Chapter 8 Section 1 to 5 author: Mr. Park
Strategy for evaluating ∫tanm xsecn x dx
a) If the power of secant is even (n = 2k ), use sec2 x =1+ tan2 x :
∫tanm xsec2k x dx = ∫tanm x(sec2 x)k−1sec2 x dx =∫tanm x(1+ tan2 x)k−1sec2 x dx then let u = tan x
b) If the power of tangent is odd (m = 2k +1), use tan2 x = sec2 x −1:
21kn+−2kn1 2 kn−1
tan xsec xdx==(tan x) sec xsecxtanxdx (sec x−1) sec xsecxtanxdx
∫∫ ∫
then let u = secx
Recall ∫tan x dx = lnsecx +C ∫secx dx = lnsecx+ tan x +C
To evaluate the following:
a) sinmxcosnx dx use sin AcosB = 1[sin(A− B)+sin(A+ B)]
∫ 2
b) sinmxsinnx dx use sin AsinB = 1[cos(A− B)−cos(A+ B)]
∫ 2
c) cosmxcosnx dx use cosAcosB = 1[cos(A−B)+cos(A+B)]
∫ 2
Section 8.4: Trigonometric Substitution (TRIANGULATION)
For this section, it will be easier to recall the basic trigonometry of a right triangle. Given triangle below:
Recall: SOH CAH TOA
opp v adj h opp v
sinθθ==cos ==tanθ==
r hyp r hyp r adj h
222
By Pythagorean theorem:
v rv= +h
If we solve for hypotenuse and each of the legs, we get:
22 22 22
rv=+hv=−rhh=−rv
θ The trick to this section is to recognize which of the following is present in the
h problem:
222222
vh+ r−−h rv
222222
vh+−rh r−v
22 22
If this expression ( vh+or vh+) is present then this part represents the hypotenuse of the triangle;
therefore, each part represent the legs of the triangle.
22 22 22 22
If these expressions ( rh−−or rh) or ( rv− or rv− ) are present then this part represents the
leg of the triangle; therefore, the fist part is the hypotenuse and second the other leg of the triangle.
Now use this triangle to pick out 2 trigonometric relationships that involve the pairs given below:
222222
sin θ +=cos θθ1 tan +1=sec θ1+cot θ=csc θ
This is the encoding step. Then use techniques of trigonometric integration to solve the problem. After solving,
use the triangle we set up for encoding to decode our solution.
City College of New York MATH 21200 (Calculus 2 Notes) Page 3 of 3
th
Thomas’ Calculus Early Transcendentals, 14 ed Chapter 8 Section 1 to 5 author: Mr. Park
Section 8.5: Integration of Rational Function by Partial Fractions
Px()
If fx()= such that deg(Px( )) ≥ deg(Qx( )), then use the long division to obtain
Qx()
Px() Rx()
fx()==S()x+ where and are polynomials.
Qx() Qx() Sx() R()x
Case 1: The denominator is a product of distinct linear factors.
Qx()
This means that we can write Qx()=+(ax b)(ax+b)"(ax+b) where no factor is repeated. In this case
1122nn
the partial fraction theorem states that there exist constants AA,,…,A such that
12 k
Rx() A A A
12k.
"
=+++
Qx() ax++b ax b ax+b
1122 kk
Case 2: is a product of linear factors, some which are repeated.
Qx()
(ax+b) r
Suppose the first linear factor is repeated r times; that is, (ax+b) occurs in the factorization of
11 11
A
. Then instead of the single term 1 in previous case 1, we would use
Qx() ()ax+b
11
AAA
12r.
"
+++
2 r
()()()
ax++b ax b ax+b
1122 rr
Case 3: contains irreducible quadratic factors, none of which is repeated.
Qx()
If has the factor 2 , where 2 , then, in addition to the partial fractions in equations
Qx() ax ++bx c ba−4c<0
R()x AxB+
from case 1 and 2, the expression for will have a term of the form 2 where A and B are
Qx() ax +bx+c
constants to be determined.
The term AxB+ can be integrated by completing the square and using the formula
ax2 ++bx c
dx 1 x
−1 ⎛⎞ .
C
=+
∫ 22tan ⎜⎟
xaa a
+ ⎝⎠
Case 4: contains a repeated irreducible quadratic factor.
Qx()
2 r 2
If has the factor ()ax ++bx c , where , then instead of the single partial fraction in case 3,
Qx() ba−4c<0
Ax++B Ax B Ax+B
11 22 rr
the sum ++"+ occurs in the partial fraction decomposition of
2 2 r
++ 22
ax bx c ++ ++
ax bx c ax bx c
()()
R()x
. Each of the terms above can be integrated by first completing the square.
Qx()
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