∫

∑

∂

∞

√

∏

Home/Calculus/Indefinite Integration/Integration by Parts

CALC-5.4

4-5 hours

Integration by Parts

The product rule in reverse: a powerful technique for products of functions.

Learning Objectives

Understand integration by parts as the product rule in reverse
Apply the LIATE rule for choosing u and dv
Use the tabular method for repeated integration by parts
Integrate products of polynomials with exponentials or trig
Integrate logarithmic and inverse trigonometric functions

1. The Integration by Parts Formula

Theorem 5.15: Integration by Parts

If u and v are differentiable functions, then:

\int u\,dv = uv - \int v\,du

Or equivalently, if u = f(x) and v = g(x):

\int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx

Proof:

Start from the product rule:

\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}

Integrate both sides:

uv = \int u\,dv + \int v\,du

Rearranging: $\int u\,dv = uv - \int v\,du$ ∎

∎

2. The LIATE Rule

Choose u (in order of priority):

LLogarithmic: ln x, log x

IInverse trig: arcsin x, arctan x, etc.

AAlgebraic: x, x², polynomials

TTrigonometric: sin x, cos x, etc.

EExponential: eˣ, 2ˣ, etc.

Choose u from higher in the list, dv from lower.

Example 5.43: ∫x eˣ dx

Problem: Evaluate $\int x e^x\,dx$

Solution:

By LIATE: u = x (Algebraic), dv = eˣ dx (Exponential)

Then: du = dx, v = eˣ

\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1) + C

Example 5.44: ∫ln x dx

Problem: Evaluate $\int \ln x\,dx$

Solution:

Let u = ln x, dv = dx. Then du = dx/x, v = x

\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C

Example 5.45: ∫x cos x dx

Problem: Evaluate $\int x \cos x\,dx$

Solution:

Let u = x, dv = cos x dx. Then du = dx, v = sin x

\int x \cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C

3. Repeated Integration by Parts

Example 5.46: ∫x² eˣ dx

Problem: Evaluate $\int x^2 e^x\,dx$

Solution (First application):

u = x², dv = eˣ dx → du = 2x dx, v = eˣ

\int x^2 e^x\,dx = x^2 e^x - \int 2x e^x\,dx

Second application:

u = 2x, dv = eˣ dx → du = 2 dx, v = eˣ

= x^2 e^x - \left(2xe^x - \int 2e^x\,dx\right) = x^2 e^x - 2xe^x + 2e^x + C

= e^x(x^2 - 2x + 2) + C

Remark 5.8: Tabular Method

For integrals of the form $\int x^n e^{ax}\,dx$ or $\int x^n \sin(ax)\,dx$ , use the tabular method:

Create two columns: derivatives of u (until 0) and integrals of dv

Multiply diagonally with alternating signs: +, -, +, -, ...

4. Cycling Integrals

Example 5.47: ∫eˣ sin x dx

Problem: Evaluate $\int e^x \sin x\,dx$

Solution:

Let I = $\int e^x \sin x\,dx$

First application: u = sin x, dv = eˣ dx

I = e^x \sin x - \int e^x \cos x\,dx

Second application: u = cos x, dv = eˣ dx

I = e^x \sin x - \left(e^x \cos x - \int e^x (-\sin x)\,dx\right)

I = e^x \sin x - e^x \cos x - \int e^x \sin x\,dx = e^x(\sin x - \cos x) - I

Solve for I:

2I = e^x(\sin x - \cos x) \Rightarrow I = \frac{e^x(\sin x - \cos x)}{2} + C

5. Inverse Trigonometric Functions

Example 5.48: ∫arctan x dx

Problem: Evaluate $\int \arctan x\,dx$

Solution:

Let u = arctan x, dv = dx. Then du = dx/(1+x²), v = x

\int \arctan x\,dx = x\arctan x - \int \frac{x}{1+x^2}\,dx

For the remaining integral, let w = 1+x², dw = 2x dx:

= x\arctan x - \frac{1}{2}\ln(1+x^2) + C

Example 5.49: ∫arcsin x dx

Problem: Evaluate $\int \arcsin x\,dx$

Solution:

Let u = arcsin x, dv = dx. Then du = dx/√(1-x²), v = x

\int \arcsin x\,dx = x\arcsin x - \int \frac{x}{\sqrt{1-x^2}}\,dx

For the remaining integral, let w = 1-x², dw = -2x dx:

= x\arcsin x + \sqrt{1-x^2} + C

Integration by Parts Quiz

Questions

Correct

Accuracy

\int x e^x\,dx =

Easy

Not attempted

\int \ln x\,dx =

Easy

Not attempted

\int x^2 e^x\,dx

requires how many applications?

Medium

Not attempted

In LIATE, what does L stand for?

Easy

Not attempted

\int x \cos x\,dx =

Medium

Not attempted

\int e^x \sin x\,dx

requires:

Hard

Not attempted

\int \arctan x\,dx =

Hard

Not attempted

For

\int x^3 e^{2x}\,dx

, which method is efficient?

Medium

Not attempted

Frequently Asked Questions

What is the integration by parts formula?

∫u dv = uv - ∫v du. It comes from the product rule: d(uv) = u dv + v du.

What is the LIATE rule?

A priority guide for choosing u: Logarithmic > Inverse trig > Algebraic > Trig > Exponential. Choose u from higher in the list.

When does integration by parts cycle?

For integrals like ∫eˣ sin x dx. After two applications, the original integral returns. Set up equation I = ... - I and solve.

What is the tabular method?

A shortcut for repeated parts with polynomial times exponential/trig. Create columns of derivatives and integrals, multiply diagonally with alternating signs.

How do I integrate ln x?

Let u = ln x, dv = dx. Then du = dx/x, v = x. Result: x ln x - x + C.

Can I choose u and dv differently?

Yes, but a bad choice may make the integral harder. LIATE usually gives the best choice.

6. Common Mistakes

Wrong choice of u

Following LIATE usually gives the simplest result. Bad choices make the integral harder.

Sign errors

Remember: ∫u dv = uv - ∫v du. The minus sign is crucial.

Not recognizing cycling

When the original integral reappears, solve algebraically for I.

7. More Worked Examples

Example 5.50: x² sin x

Problem: $\int x^2 \sin x\,dx$

First application: u = x², dv = sin x dx

= -x^2\cos x + 2\int x\cos x\,dx

Second application: u = x, dv = cos x dx

= -x^2\cos x + 2(x\sin x + \cos x) + C = -x^2\cos x + 2x\sin x + 2\cos x + C

Example 5.51: x³ eˣ

Problem: $\int x^3 e^x\,dx$

Using tabular method (3 applications):

= x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x + C = e^x(x^3 - 3x^2 + 6x - 6) + C

Example 5.52: eˣ cos x

Problem: $\int e^x \cos x\,dx$

Let I = ∫eˣ cos x dx. Apply parts twice:

I = e^x\cos x + e^x\sin x - I

2I = e^x(\cos x + \sin x) \Rightarrow I = \frac{e^x(\cos x + \sin x)}{2} + C

Example 5.53: ln² x

Problem: $\int (\ln x)^2\,dx$

Let u = (ln x)², dv = dx:

= x(\ln x)^2 - 2\int \ln x\,dx = x(\ln x)^2 - 2(x\ln x - x) + C

= x(\ln x)^2 - 2x\ln x + 2x + C

Example 5.54: x arcsin x

Problem: $\int x \arcsin x\,dx$

Let u = arcsin x, dv = x dx:

= \frac{x^2}{2}\arcsin x - \int \frac{x^2}{2\sqrt{1-x^2}}\,dx

The remaining integral uses substitution:

= \frac{x^2}{2}\arcsin x + \frac{1}{2}\sqrt{1-x^2} - \frac{1}{2}\arcsin x + C

8. Tabular Method Explained

Steps:

Create two columns: derivatives of u (until 0) and integrals of dv
Draw diagonal arrows connecting adjacent rows
Alternate signs: +, -, +, -, ...
Multiply along each diagonal and sum

Example 5.55: Tabular: x⁴eˣ

Sign	u (diff)	dv (int)
+	x⁴	eˣ
−	4x³	eˣ
+	12x²	eˣ
−	24x	eˣ
+	24	eˣ
	0	eˣ

= e^x(x^4 - 4x^3 + 12x^2 - 24x + 24) + C

9. Practice Problems

1. $\int x \sin 2x\,dx$

Answer

$-\frac{x}{2}\cos 2x + \frac{1}{4}\sin 2x + C$

2. $\int x^2 \ln x\,dx$

Answer

$\frac{x^3}{3}\ln x - \frac{x^3}{9} + C$

3. $\int e^{2x} \sin x\,dx$

Answer

$\frac{e^{2x}(2\sin x - \cos x)}{5} + C$

4. $\int \arccos x\,dx$

Answer

$x\arccos x - \sqrt{1-x^2} + C$

5. $\int x e^{-x}\,dx$

Answer

$-xe^{-x} - e^{-x} + C = -e^{-x}(x+1) + C$

6. $\int x^2 \cos x\,dx$

Answer

$x^2\sin x + 2x\cos x - 2\sin x + C$

10. Special Cases

Example 5.56: sec³x

Problem: $\int \sec^3 x\,dx$

Let u = sec x, dv = sec²x dx:

I = \sec x \tan x - \int \sec x \tan^2 x\,dx

Use tan²x = sec²x - 1:

I = \sec x \tan x - \int (\sec^3 x - \sec x)\,dx = \sec x\tan x - I + \ln|\sec x + \tan x|

2I = \sec x\tan x + \ln|\sec x + \tan x| \Rightarrow I = \frac{1}{2}(\sec x\tan x + \ln|\sec x + \tan x|) + C

Example 5.57: √(a²+x²)

Problem: $\int \sqrt{a^2+x^2}\,dx$

Let u = √(a²+x²), dv = dx:

= x\sqrt{a^2+x^2} - \int \frac{x^2}{\sqrt{a^2+x^2}}\,dx

Simplify the remaining integral:

= x\sqrt{a^2+x^2} - \int \sqrt{a^2+x^2}\,dx + a^2\int \frac{dx}{\sqrt{a^2+x^2}}

2I = x\sqrt{a^2+x^2} + a^2\ln|x + \sqrt{a^2+x^2}|

I = \frac{x\sqrt{a^2+x^2}}{2} + \frac{a^2}{2}\ln|x + \sqrt{a^2+x^2}| + C

11. Study Tips

1. LIATE is a Guide

Not an absolute rule, but works well in most cases.

2. Check Your Choice

If the integral gets harder, try switching u and dv.

3. Watch for Patterns

polynomial × exp/trig → tabular method

4. Cycling Integrals

eˣ sin x, eˣ cos x require solving for I

5. Verify

Always differentiate to check your answer.

6. Combine Methods

May need substitution after parts or vice versa.

12. Additional Examples

∫x ln²x dx

= \frac{x^2}{2}(\ln x)^2 - \frac{x^2}{2}\ln x + \frac{x^2}{4} + C

∫x² arctan x dx

= \frac{x^3}{3}\arctan x - \frac{x^2}{6} + \frac{1}{6}\ln(1+x^2) + C

∫sin x ln(cos x) dx

= \cos x - \cos x \ln(\cos x) + C

13. Common Integration by Parts Results

Integral	Result
$\int \ln x\,dx$	$x\ln x - x + C$
$\int x e^x\,dx$	$e^x(x-1) + C$
$\int x\sin x\,dx$	$-x\cos x + \sin x + C$
$\int x\cos x\,dx$	$x\sin x + \cos x + C$
$\int \arctan x\,dx$	$x\arctan x - \frac{1}{2}\ln(1+x^2) + C$
$\int \arcsin x\,dx$	$x\arcsin x + \sqrt{1-x^2} + C$
$\int e^x\sin x\,dx$	$\frac{e^x(\sin x - \cos x)}{2} + C$

14. Final Practice Set

P1: $\int x^3 \sin x\,dx$

Answer

$-x^3\cos x + 3x^2\sin x + 6x\cos x - 6\sin x + C$

P2: $\int x^2 e^{-x}\,dx$

Answer

$-e^{-x}(x^2 + 2x + 2) + C$

P3: $\int e^{-x} \cos x\,dx$

Answer

$\frac{e^{-x}(\sin x - \cos x)}{2} + C$

P4: $\int x \arctan x\,dx$

Answer

$\frac{x^2}{2}\arctan x - \frac{x}{2} + \frac{1}{2}\arctan x + C$

15. Advanced Examples

Example 5.58: ∫x²sin²x dx

Use identity sin²x = (1-cos2x)/2 first:

\int x^2 \sin^2 x\,dx = \frac{1}{2}\int x^2(1-\cos 2x)\,dx

= \frac{x^3}{6} - \frac{1}{2}\int x^2\cos 2x\,dx

Apply tabular method for the second integral.

Example 5.59: ∫eˣ sin²x dx

Use sin²x = (1-cos2x)/2:

\int e^x \sin^2 x\,dx = \frac{1}{2}\int e^x\,dx - \frac{1}{2}\int e^x\cos 2x\,dx

= \frac{e^x}{2} - \frac{e^x(2\sin 2x + \cos 2x)}{10} + C

Example 5.60: ∫ln(1+x²) dx

Let u = ln(1+x²), dv = dx:

= x\ln(1+x^2) - \int \frac{2x^2}{1+x^2}\,dx

= x\ln(1+x^2) - 2x + 2\arctan x + C

16. Reduction Formulas

∫sinⁿx dx

$I_n = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}I_{n-2}$

∫cosⁿx dx

$I_n = \frac{\cos^{n-1}x\sin x}{n} + \frac{n-1}{n}I_{n-2}$

∫secⁿx dx

$I_n = \frac{\sec^{n-2}x\tan x}{n-1} + \frac{n-2}{n-1}I_{n-2}$

∫(ln x)ⁿ dx

$I_n = x(\ln x)^n - nI_{n-1}$

17. More Practice Problems

Q1: $\int x \sin^2 x\,dx$

Answer

$\frac{x^2}{4} - \frac{x\sin 2x}{4} - \frac{\cos 2x}{8} + C$

Q2: $\int (\ln x)^3\,dx$

Answer

$x(\ln x)^3 - 3x(\ln x)^2 + 6x\ln x - 6x + C$

Q3: $\int x^4 e^{2x}\,dx$

Answer

$\frac{e^{2x}}{2}(x^4 - 2x^3 + 3x^2 - 3x + \frac{3}{2}) + C$

Q4: $\int e^{ax} \sin bx\,dx$

Answer

$\frac{e^{ax}(a\sin bx - b\cos bx)}{a^2+b^2} + C$

Q5: $\int \sec^5 x\,dx$

Hint

Use reduction formula: I₅ = (sec³x tan x)/4 + (3/4)I₃

18. Chapter Summary

When to Use

• Product of different function types
• Single inverse trig or log functions
• Powers of trig with exp

Key Techniques

• LIATE rule for choosing u
• Tabular method for polynomials
• Solve for I when cycling

19. Quick Reference Card

Log/Inv Trig

• u = ln x, arctan x, etc.
• dv = polynomial dx

Poly × Exp/Trig

• u = polynomial
• dv = eˣ dx or trig dx

Exp × Trig

• Apply twice
• Solve for I

20. Additional Practice

$\int x^2 \arctan x\,dx = \frac{x^3}{3}\arctan x - \frac{x^2}{6} + \frac{\ln(1+x^2)}{6} + C$

$\int x \sec^2 x\,dx = x\tan x + \ln|\cos x| + C$

$\int e^{2x} \cos 3x\,dx = \frac{e^{2x}(2\cos 3x + 3\sin 3x)}{13} + C$

$\int \ln x \sin x\,dx = -\ln x \cos x + \text{Ci}(x) + C$

21. Even More Examples

Example 5.61: x arctan²x

Problem: $\int x (\arctan x)^2\,dx$

Let u = (arctan x)², dv = x dx:

= \frac{x^2}{2}(\arctan x)^2 - \int \frac{x^2 \arctan x}{1+x^2}\,dx

Example 5.62: x² ln²x

Problem: $\int x^2 (\ln x)^2\,dx$

Apply parts twice:

= \frac{x^3}{3}(\ln x)^2 - \frac{2x^3}{9}\ln x + \frac{2x^3}{27} + C

Example 5.63: eˣ sin x cos x

Problem: $\int e^x \sin x \cos x\,dx$

Use sin x cos x = (sin 2x)/2:

= \frac{1}{2}\int e^x \sin 2x\,dx = \frac{e^x(\sin 2x - 2\cos 2x)}{10} + C

22. Challenge Problems

C1: $\int x^5 e^{x^2}\,dx$

Hint

First substitute u = x², then use parts

C2: $\int (\ln x)^4\,dx$

Hint

Use reduction: Iₙ = x(ln x)ⁿ - nIₙ₋₁

C3: $\int e^{\sqrt{x}}\,dx$

Hint

Substitute u = √x first, then parts

23. Final Summary

Type 1: ln x, arctan x

u = log/inv trig, dv = algebraic

Type 2: xⁿ × eˣ/sin/cos

u = polynomial, use tabular method

Type 3: eˣ × sin/cos

Apply twice, solve for I

24. Additional Examples

$\int x \ln(x+1)\,dx = \frac{(x^2-1)\ln(x+1)}{2} - \frac{x^2}{4} + \frac{x}{2} + C$

$\int x^2 e^{3x}\,dx = \frac{e^{3x}(9x^2-6x+2)}{27} + C$

$\int x \cosh x\,dx = x\sinh x - \cosh x + C$

$\int \arcsin^2 x\,dx = x\arcsin^2 x + 2\sqrt{1-x^2}\arcsin x - 2x + C$

25. More Worked Examples

Example 5.64: x sinh x

Problem: $\int x \sinh x\,dx$

Let u = x, dv = sinh x dx:

= x\cosh x - \int \cosh x\,dx = x\cosh x - \sinh x + C

Example 5.65: x² arcsin x

Problem: $\int x^2 \arcsin x\,dx$

= \frac{x^3}{3}\arcsin x - \int \frac{x^3}{3\sqrt{1-x^2}}\,dx

The remaining integral uses substitution u = 1-x²

Example 5.66: ln(ln x)

Problem: $\int \ln(\ln x)\,dx$

Let u = ln(ln x), dv = dx:

= x\ln(\ln x) - \int \frac{1}{\ln x}\,dx

The integral ∫1/ln x dx = li(x) (logarithmic integral)

26. Final Practice Set

F1: $\int x^5 \cos x\,dx$

Hint

Use tabular method (5 rows)

F2: $\int e^{-x} \sin 2x\,dx$

Answer

$\frac{e^{-x}(-\sin 2x - 2\cos 2x)}{5} + C$

F3: $\int x \text{arcsec} x\,dx$

Hint

u = arcsec x, dv = x dx

F4: $\int (\ln x)^5\,dx$

Hint

Reduction: Iₙ = x(ln x)ⁿ - nIₙ₋₁

27. Important Results Summary

\int x^n e^x\,dx = e^x \sum_{k=0}^{n}(-1)^{n-k}\frac{n!}{k!}x^k + C

\int e^{ax}\sin bx\,dx = \frac{e^{ax}(a\sin bx - b\cos bx)}{a^2+b^2} + C

\int e^{ax}\cos bx\,dx = \frac{e^{ax}(a\cos bx + b\sin bx)}{a^2+b^2} + C

28. Additional Practice

$\int x^2 \sin 2x\,dx$

Answer

$-\frac{x^2\cos 2x}{2} + \frac{x\sin 2x}{2} + \frac{\cos 2x}{4} + C$

$\int x^3 \ln^2 x\,dx$

Answer

$\frac{x^4}{4}\ln^2 x - \frac{x^4}{8}\ln x + \frac{x^4}{32} + C$

$\int e^{3x} \cos 4x\,dx$

Answer

$\frac{e^{3x}(3\cos 4x + 4\sin 4x)}{25} + C$

$\int x \text{arctan}^2 x\,dx$

Hint

u = arctan²x, dv = x dx

29. Final Worked Examples

Example 5.67: x²e²ˣ

Using tabular method:

\int x^2 e^{2x}\,dx = \frac{e^{2x}}{4}(2x^2 - 2x + 1) + C

Example 5.68: x arctan(x²)

Let u = arctan(x²), dv = x dx:

= \frac{x^2}{2}\arctan(x^2) - \frac{1}{2}\int \frac{x^3}{1+x^4}\,dx

= \frac{x^2}{2}\arctan(x^2) - \frac{1}{8}\ln(1+x^4) + C

30. Chapter Summary

Formula

$\int u\,dv = uv - \int v\,du$

LIATE Rule

Log → Inv Trig → Algebraic → Trig → Exp

Methods

Tabular, Cycling, Reduction

31. Even More Examples

∫x cosh x dx

= x\sinh x - \cosh x + C

∫x² ln(x²) dx

= \frac{x^3}{3}\ln(x^2) - \frac{2x^3}{9} + C = \frac{2x^3}{3}\ln x - \frac{2x^3}{9} + C

∫eˣ x² sin x dx

Requires multiple applications

∫arctan²x dx

= x\arctan^2 x - \ln(1+x^2)\arctan x + x\ln(1+x^2) - 2x + 2\arctan x + C

32. Final Practice Set

$\int x^4 \sin x\,dx$

Method

Use tabular method (4 applications)

$\int (\ln x)^4\,dx$

Method

Reduction: I₄ = x(ln x)⁴ - 4I₃

$\int e^{-2x} \sin 3x\,dx$

Answer

$\frac{e^{-2x}(-2\sin 3x - 3\cos 3x)}{13} + C$

$\int x^3 e^{-x}\,dx$

Answer

$-e^{-x}(x^3 + 3x^2 + 6x + 6) + C$

33. Additional Results

\int x^n \ln x\,dx = \frac{x^{n+1}}{n+1}\ln x - \frac{x^{n+1}}{(n+1)^2} + C

\int x^n e^{ax}\,dx = \frac{x^n e^{ax}}{a} - \frac{n}{a}\int x^{n-1}e^{ax}\,dx

\int x^n \sin ax\,dx = -\frac{x^n\cos ax}{a} + \frac{n}{a}\int x^{n-1}\cos ax\,dx

\int x^n \cos ax\,dx = \frac{x^n\sin ax}{a} - \frac{n}{a}\int x^{n-1}\sin ax\,dx

34. When to Use Integration by Parts

Integrand Type	u Choice	dv Choice
xⁿ eˣ	xⁿ	eˣ dx
xⁿ sin x / cos x	xⁿ	sin/cos dx
xⁿ ln x	ln x	xⁿ dx
xⁿ arctan/arcsin	arctan/arcsin	xⁿ dx
eˣ sin x / cos x	eˣ or sin/cos	(cycling)

35. Final Quick Reference

LIATE Priority:

L - Logarithmic | I - Inverse trig | A - Algebraic | T - Trig | E - Exponential

Cycling Integrals:

Apply parts twice, solve resulting equation for I

Key Takeaways

Formula

$\int u\,dv = uv - \int v\,du$

LIATE

Log, Inverse trig, Algebraic, Trig, Exponential

Tabular Method

For polynomial × exp/trig - derivatives & integrals in columns

Cycling

When I reappears: solve 2I = ... for I

Follow LIATE for choosing u
Use tabular method for polynomials
Watch for cycling integrals
Always verify by differentiating