Integral Forms and Trigonometric Integrals MATH 166 Fall 2011 Integral forms that you need to know Let k be a real number, r a rational number and a> 0 be a positive real number. 1. k du = ku + C 2. u r du = ⎧ ⎪ ⎨ ⎪ ⎩ u r+1 r +1 + C, r ̸= −1 ln |u| + C, r = −1 3. e u du = e u + C 4. a u du = a u ln a + C, a ̸=1 5. sin u du = − cos u + C 6. cos u du = sin u + C 7. sec 2 u du = tan u + C 8. csc 2 u du = − cot u + C 9. sec u tan u du = sec u + C 10. csc u cot u du = − csc u + C 11. tan u du = − ln | cos u| + C 12. cot u du = ln | sin u| + C 13. du √ a 2 − u 2 = sin −1 u a + C 14. du a 2 + u 2 = 1 a tan −1 u a + C 15. du u √ u 2 − a 2 = 1 a sec −1 |u| a + C 16. sinh u du = cosh u + C 17. cosh u du = sinh u + C Taken from pp. 383-384 in Calculus, 9th ed. by Varberg, Purcell and Rigdon. 1
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Integral Forms and Trigonometric Integrals
MATH 166
Fall 2011
Integral forms that you need to know
Let k be a real number, r a rational number and a > 0 be a positive real number.
1.
∫
k du = ku+ C
2.
∫
ur du =
⎧
⎪
⎨
⎪
⎩
ur+1
r + 1+ C, r ̸= −1
ln |u|+ C, r = −1
3.
∫
eu du = eu + C
4.
∫
au du =au
ln a+ C, a ̸= 1
5.
∫
sin u du = − cosu+ C
6.
∫
cosu du = sin u+ C
7.
∫
sec2 u du = tanu+ C
8.
∫
csc2 u du = − cot u+ C
9.
∫
sec u tanu du = sec u+ C
10.
∫
csc u cotu du = − csc u+ C
11.
∫
tan u du = − ln | cosu|+ C
12.
∫
cot u du = ln | sin u|+ C
13.
∫
du√a2 − u2
= sin−1
(u
a
)
+ C
14.
∫
du
a2 + u2=
1
atan−1
(u
a
)
+ C
15.
∫
du
u√u2 − a2
=1
asec−1
(
|u|a
)
+ C
16.
∫
sinh u du = cosh u+ C
17.
∫
cosh u du = sinh u+ C
Taken from pp. 383-384 in Calculus, 9th ed. by Varberg, Purcell and Rigdon.
1
Integration by parts∫
u dv = uv −∫
v du
∫ b
au dv = [uv]ba −
∫ b
av du.
Look for a product of functions such that the derivative of one and the inte-gral of another is a basic form.
Simple radicals
Look for n
√ax+ b:
u = (ax+ b)1/n
un = ax+ b ⇔ x = (un − b)/a
nun−1 du = a dx
Trigonometric substitution√a2 − x2 x = a sin t −π/2 ≤ t ≤ π/2√a2 + x2 x = a tan t −π/2 < t < π/2√x2 − a2 x = a sec t 0 ≤ t ≤ π, t ̸= π/2
Can drop absolute value bars.Use sin−1, tan−1, sec−1 or triangles to back-substitute.
∫
sec x dx = ln |sec x+ tanx|+ C∫
csc x dx = ln |csc x− cotx|+ C
Completing the square
x2 +Bx + C = x2 + 2(B/2)x+ (B/2)2 +[
C − (B/2)2]
= (x+ B/2)2 +[
C −B2/4]
3
Rational functions
Let p(x), q(x) be polynomials.∫
p(x)
q(x)dx
0. Do long division to ensure that the degree of p(x) is less than the degreeof q(x).
1. Factor q(x) into linear and (unfactorable) quadratic terms.
2. Each term of the factorization of q(x) gives a cluster of terms.
• The term (x− a)n gives rise to the cluster of terms:
A1
x− a+
A2
(x− a)2+ · · ·+
An
(x− a)n.
• The term (x2 + bx+ c)m gives rise to the cluster of terms:
B1x+ C1
x2 + bx+ c+
B2x+ C2
(x2 + bx+ c)2+ · · ·+
Bmx+ Cm
(x2 + bx+ c)m.
3. Solve for the unknown coefficients to get a Partial Fraction Decomposi-tion.
4. Integrate using∫
dx
x− a= ln |x− a| + C
∫
2x+ b
x2 + bx+ cdx = ln |x2 + bx+ c|+ C
∫
dx
(x+ d)2 + a2=
1
atan−1
(
x+ d
a
)
+ C.
4
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