Complete Integration Formulas: Basic to Intermediate
This guide covers every essential integration formula from fundamentals to intermediate techniques, with worked examples and practice problems.
1. Basic Integration Rules
1.1 Fundamental Rules
| Rule | Formula |
|---|
| Constant Rule | ∫kdx=kx+C |
| Power Rule | ∫xndx=n+1x, where |
| Constant Multiple | ∫kf(x)dx=k∫f(x)dx |
| Sum/Difference | ∫[f(x)±g(x)]dx=∫f(x)dx |
1.2 Example: Power Rule
Problem: Evaluate ∫3x4dx
Solution:
∫3x4dx=3⋅
2. Standard Algebraic Integrals
∫dx=x+C
∫xndx=n+1x
∫x1dx=ln∣x∣+C
∫exdx=ex+C
∫axdx=lnaa
2.1 Example: Exponential and Logarithmic
Problem: Evaluate ∫x5dx+∫2exdx
Solution:
∫x5dx+∫2e
3. Trigonometric Integrals
3.1 Basic Trigonometric Formulas
∫sinxdx=−cosx+C
∫cosxdx=sinx+C
∫tanxdx=−ln∣cosx∣+C=ln∣secx
∫cotxdx=ln∣sinx∣+C
∫secxdx=ln∣secx+tanx∣+C
∫cscxdx=ln∣cscx−cotx∣+C=
3.2 Square Terms
∫sec2xdx=tanx+C
∫csc2xdx=−cotx+C
∫secxtanxdx=secx+C
∫cscxcotxdx=−cscx+C
3.3 Example: Trigonometric Integration
Problem: Evaluate ∫(2sinx+3cosx)dx
Solution:
∫(2sinx+3cosx)dx=−2cosx+3sinx
4. Inverse Trigonometric Integrals
∫1−x2
∫1−x2
∫1+x21dx=tan
∫1+x2−1dx=cot
∫xx2−1
∫xx2−1
4.1 Generalized Forms
∫a2−x2
∫a2+x21
∫xx2−a
4.2 Example: Inverse Trigonometric
Problem: Evaluate ∫9+x21dx
Solution:
Here a2=9, so a=3:
∫9+x21dx=
5. Hyperbolic Integrals
∫sinhxdx=coshx+C
∫coshxdx=sinhx+C
∫tanhxdx=ln(coshx)+C
∫sech2xdx=tanhx+C
∫csch2xdx=−cothx+C
6. Integration by Substitution
6.1 Method
If an integral has the form ∫f(g(x))⋅g′(x)dx, substitute u, so .
6.2 Common Substitutions
| Integral Form | Substitution |
|---|
| ∫f(ax+b)dx | u=ax+b |
|
6.3 Example 1: Linear Substitution
Problem: Evaluate ∫(3x+2)5dx
Solution:
Let u=3x+2, then du=3dx, so dx=.
∫(3x+2)5dx=∫
6.4 Example 2: Logarithmic Type
Problem: Evaluate ∫x2+12xdx
Solution:
Let u=x2+1, then du=2xdx.
∫x2+12x
6.5 Example 3: Trigonometric Substitution
Problem: Evaluate ∫sin3xcosxdx
Solution:
Let u=sinx, then du=cosxdx.
∫sin3xcosxdx=∫u
7. Integration by Parts
7.1 Formula
∫udv=uv−∫vdu
Or equivalently:
∫f(x)g′(x)dx=f(x)g
7.2 LIATE Rule (Priority for choosing u)
Choose u in this order of priority:
- L — Logarithmic functions (lnx)
- I — Inverse trigonometric (sin−1x, tan−1x)
7.3 Example 1: Algebraic × Exponential
Problem: Evaluate ∫xexdx
Solution:
Let u=x (algebraic, priority over exponential) and dv=exdx.
Then du=dx and v=ex.
∫xexdx=xex−
7.4 Example 2: Logarithmic
Problem: Evaluate ∫lnxdx
Solution:
Let u=lnx and dv=dx.
Then du=x1dx and v=x.
∫lnxdx=xlnx−∫x⋅
7.5 Example 3: Repeated Integration by Parts
Problem: Evaluate ∫x2sinxdx
Solution:
First application: u=x2, dv=sinxdx
- du=2xdx, v=−cosx
∫x2sinxdx=−x2cosx+
Second application for ∫2xcosxdx:
- u=2x, dv=cosxdx
- du=2dx,
∫2xcosxdx=2xsinx−∫2sinxdx
Final answer:
∫x2sinxdx=−x2cosx+
8. Integration of Rational Functions (Partial Fractions)
8.1 Types of Partial Fractions
| Form of Factor in Denominator | Form of Partial Fraction |
|---|
| Non-repeated linear: (ax+b) | ax+bA |
8.2 Example 1: Distinct Linear Factors
Problem: Evaluate ∫(x−1)(x+2)3x+5dx
Solution:
(x−1)(x+2)3x+5=
Multiply by (x−1)(x+2):
3x+5=A(x+2)+B(x−1)
- Set x=1: 8=3A⇒A=38
∫(x−1)(x+2)3x+
8.3 Example 2: Repeated Linear Factor
Problem: Evaluate ∫(x−1)2x+1dx
Solution:
(x−1)2x+1=
Multiply by (x−1)2:
x+1=A(x−1)+B
- Set x=1: 2=B
- Compare x coefficients: 1=A, so
∫(x−1)2
9. Trigonometric Integrals (Advanced)
9.1 Powers of Sine and Cosine
Case 1: ∫sinmxcosnxdx where m is odd.
- Save one sinx, convert rest to cosx using sin2x=1−cos.
Case 2: n is odd.
- Save one cosx, convert rest to sinx using cos2x=1−sin.
Case 3: Both m and n are even.
sin2x=21−cos2x
9.2 Example: Even Powers
Problem: Evaluate ∫sin2xdx
Solution:
∫sin2xdx=∫
9.3 Products to Sums
∫sin(mx)cos(nx)dx,∫sin(mx
Use identities:
sinAcosB=21[sin(A+B)
sinAsinB=21[cos(A−B)
cosAcosB=21[cos(A+B)
10. Trigonometric Substitution
10.1 Standard Substitutions
| Expression | Substitution | Identity Used |
|---|
| a2−x2 |
10.2 Example: a2−x2 Form
Problem: Evaluate ∫4−x2
Solution:
Let x=2sinθ, then dx=2cosθdθ.
4−x2=
∫4−x
11. Special Algebraic Integrals
11.1 Forms with Square Roots
∫x2+
∫x2−
∫x2+a2
∫x2−a2
∫a2−x2
11.2 Example: Special Form
Problem: Evaluate ∫x2+9dx
Solution:
Using the formula with a=3:
∫x2+9
12. Definite Integrals
12.1 Fundamental Theorem of Calculus
If F′(x)=f(x), then:
∫abf(x)dx=F(b)−
12.2 Properties
∫abf(x)dx=−∫
∫aaf(x)dx=0
∫abf(x)dx=∫
∫abkf(x)dx=k∫
12.3 Even and Odd Functions
If f is even (f(−x)=f(x)):
∫−aaf(x)dx=2∫
If f is odd (f(−x)=−f(x)):
∫−aaf(x)dx=0
12.4 Example: Definite Integral
Problem: Evaluate ∫02πsinxdx
Solution:
∫02
13. Reduction Formulas
13.1 Wallis' Integrals
For In=∫0:
In=nn−1I
With base cases:
- I0=2π
Final Formulas:
If n is even:
∫02π
If n is odd:
∫02π
13.2 Reduction Formula for tannx
In=∫tannxdx=
13.3 Reduction Formula for secnx
In=∫secnx
14. More Standard Integrals (Quick Reference)
14.1 Tangent Half-Angle (Weierstrass Substitution)
If t=tan2x, then:
sinx=1+t2
14.2 Special Rational Forms
∫x2−a2
∫a2−x2
∫x2+a21
∫x2+a2x
∫x2−a2x
15. Solved Examples (Mixed Practice)
Example 1: Substitution with Root
Problem: ∫x+1x
Solution:
Let u=x+1, so x=u−1 and dx=du.
∫u
Substitute back:
=32(x
Example 2: Integration by Parts — Cyclic
Problem: ∫exsinxdx
Solution:
Let I=∫exsinxdx.
First parts: u=sinx, dv=exdx
- du=cosxdx, v=ex
I=exsinx−∫excosxdx
Second parts for remaining integral: u=cosx, dv=exdx
- du=−sinxdx, v=ex
∫excosxdx=excosx
Substitute back:
I=exsinx−(excosx
2I=ex(sinx−cosx)
I=2ex(sinx−cosx)+
Example 3: Completing the Square
Problem: ∫x2+4x+51dx
Solution:
Complete the square:
x2+4x+5=(x+2)2+
Let u=x+2, du=dx:
∫u2+11du
Example 4: Definite with Substitution
Problem: ∫04x
Solution:
Let u=x, so x=u and .
When , ; when , .
∫02u
=2[u−ln∣u+1∣]0
16. Practice Problems
Level 1 — Basic
- ∫(5x3−2x+7)dx
- ∫
Level 2 — Substitution
- ∫x(x2+1)4dx
Level 3 — By Parts & Partial Fractions
- ∫xlnxdx
- ∫x2exd
Level 4 — Trigonometric & Special
- ∫sin2xcos2xdx
- ∫
17. Answers to Practice Problems
| Problem | Answer |
|---|
| 1 | 45x4−x |
Quick Reference Formula Sheet
| Integral | Result |
|---|
| ∫xndx | n+1 |
End of Integration Formulas Guide