MathIsimple
π
θ
Δ
ε
CALC-5.3
4-5 hours

Integration by Substitution

The chain rule in reverse: master u-substitution to evaluate complex integrals.

Learning Objectives
  • Understand the substitution method as the reverse of the chain rule
  • Identify suitable substitutions for various integrals
  • Execute u-substitution systematically
  • Apply trigonometric substitutions
  • Recognize patterns that suggest specific substitutions

1. Introduction to Substitution

The substitution method (also called u-substitution or change of variables) is the integration counterpart to the chain rule for differentiation. If we know that:

ddx[F(g(x))]=F(g(x))g(x)\frac{d}{dx}[F(g(x))] = F'(g(x)) \cdot g'(x)

Then reversing this:

F(g(x))g(x)dx=F(g(x))+C\int F'(g(x)) \cdot g'(x)\,dx = F(g(x)) + C

2. The Substitution Method

Theorem 5.13: Substitution Rule

If u=g(x)u = g(x) is a differentiable function and ff is continuous, then:

f(g(x))g(x)dx=f(u)du\int f(g(x)) \cdot g'(x)\,dx = \int f(u)\,du

where we substitute back u=g(x)u = g(x) after integration.

Remark 5.7: Step-by-Step Process
  1. Choose u: Let u = g(x), an “inner function”
  2. Find du: Compute du = g'(x) dx
  3. Substitute: Replace all x expressions with u expressions
  4. Integrate: Evaluate ∫f(u) du
  5. Back-substitute: Replace u with g(x)
Example 5.36: Basic u-Substitution

Problem: Evaluate 2x(x2+1)5dx\int 2x(x^2 + 1)^5\,dx

Solution:

Let u=x2+1u = x^2 + 1, then du=2xdxdu = 2x\,dx

2x(x2+1)5dx=u5du=u66+C=(x2+1)66+C\int 2x(x^2 + 1)^5\,dx = \int u^5\,du = \frac{u^6}{6} + C = \frac{(x^2+1)^6}{6} + C
Example 5.37: Exponential Composite

Problem: Evaluate esinxcosxdx\int e^{\sin x} \cos x\,dx

Solution:

Let u=sinxu = \sin x, then du=cosxdxdu = \cos x\,dx

esinxcosxdx=eudu=eu+C=esinx+C\int e^{\sin x} \cos x\,dx = \int e^u\,du = e^u + C = e^{\sin x} + C
Example 5.38: Adjusting Constants

Problem: Evaluate xex2dx\int x e^{x^2}\,dx

Solution:

Let u=x2u = x^2, then du=2xdxdu = 2x\,dx, so xdx=12dux\,dx = \frac{1}{2}du

xex2dx=12eudu=12eu+C=12ex2+C\int x e^{x^2}\,dx = \frac{1}{2}\int e^u\,du = \frac{1}{2}e^u + C = \frac{1}{2}e^{x^2} + C

3. Common Substitution Patterns

Pattern 1: Power of a linear function

(ax+b)ndx\int (ax + b)^n\,dx → Let u = ax + b

=(ax+b)n+1a(n+1)+C= \frac{(ax+b)^{n+1}}{a(n+1)} + C

Pattern 2: Function times its derivative

f(x)nf(x)dx\int f(x)^n \cdot f'(x)\,dx → Let u = f(x)

=f(x)n+1n+1+C= \frac{f(x)^{n+1}}{n+1} + C

Pattern 3: Exponential of a function

ef(x)f(x)dx\int e^{f(x)} \cdot f'(x)\,dx → Let u = f(x)

=ef(x)+C= e^{f(x)} + C

Pattern 4: Logarithmic form

f(x)f(x)dx\int \frac{f'(x)}{f(x)}\,dx → Let u = f(x)

=lnf(x)+C= \ln|f(x)| + C
Example 5.39: Logarithmic Pattern

Problem: Evaluate 2xx2+1dx\int \frac{2x}{x^2 + 1}\,dx

Solution:

Recognize: numerator = derivative of denominator

2xx2+1dx=lnx2+1+C=ln(x2+1)+C\int \frac{2x}{x^2 + 1}\,dx = \ln|x^2 + 1| + C = \ln(x^2 + 1) + C
Example 5.40: Tangent Function

Problem: Derive tanxdx\int \tan x\,dx

Solution:

Write as sinxcosx\frac{\sin x}{\cos x}. Let u = cos x, du = -sin x dx

sinxcosxdx=duu=lnu+C=lncosx+C=lnsecx+C\int \frac{\sin x}{\cos x}\,dx = -\int \frac{du}{u} = -\ln|u| + C = -\ln|\cos x| + C = \ln|\sec x| + C

4. Trigonometric Substitution

Theorem 5.14: Trigonometric Substitutions
ExpressionSubstitutionIdentity Used
a2x2\sqrt{a^2 - x^2}x=asinθx = a\sin\theta1sin2θ=cos2θ1 - \sin^2\theta = \cos^2\theta
a2+x2\sqrt{a^2 + x^2}x=atanθx = a\tan\theta1+tan2θ=sec2θ1 + \tan^2\theta = \sec^2\theta
x2a2\sqrt{x^2 - a^2}x=asecθx = a\sec\thetasec2θ1=tan2θ\sec^2\theta - 1 = \tan^2\theta
Example 5.41: Trig Substitution

Problem: Evaluate 14x2dx\int \frac{1}{\sqrt{4-x^2}}\,dx

Solution:

Let x=2sinθx = 2\sin\theta, then dx=2cosθdθdx = 2\cos\theta\,d\theta

4x2=44sin2θ=2cosθ\sqrt{4-x^2} = \sqrt{4-4\sin^2\theta} = 2\cos\theta

2cosθ2cosθdθ=dθ=θ+C=arcsinx2+C\int \frac{2\cos\theta}{2\cos\theta}\,d\theta = \int d\theta = \theta + C = \arcsin\frac{x}{2} + C
Example 5.42: Trig Substitution with √(a²+x²)

Problem: Evaluate 1x2+9dx\int \frac{1}{\sqrt{x^2+9}}\,dx

Solution:

Let x=3tanθx = 3\tan\theta, then dx=3sec2θdθdx = 3\sec^2\theta\,d\theta

x2+9=3secθ\sqrt{x^2+9} = 3\sec\theta

3sec2θ3secθdθ=secθdθ=lnsecθ+tanθ+C\int \frac{3\sec^2\theta}{3\sec\theta}\,d\theta = \int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C

Back-substitute: =lnx3+x2+93+C=lnx+x2+9+C= \ln\left|\frac{x}{3} + \frac{\sqrt{x^2+9}}{3}\right| + C = \ln|x + \sqrt{x^2+9}| + C'

Substitution Quiz
8
Questions
0
Correct
0%
Accuracy
1
2x(x2+1)5dx\int 2x(x^2+1)^5\,dx using u=x2+1u = x^2+1:
Easy
Not attempted
2
cosxesinxdx=\int \cos x \cdot e^{\sin x}\,dx =
Easy
Not attempted
3
lnxxdx=\int \frac{\ln x}{x}\,dx =
Medium
Not attempted
4
tanxdx=sinxcosxdx=\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx =
Medium
Not attempted
5
x1x2dx=\int x\sqrt{1-x^2}\,dx =
Medium
Not attempted
6
ex1+exdx=\int \frac{e^x}{1+e^x}\,dx =
Medium
Not attempted
7
For a2x2dx\int \sqrt{a^2-x^2}\,dx, what substitution?
Hard
Not attempted
8
sec3xtanxdx=\int \sec^3 x \tan x\,dx =
Medium
Not attempted

Frequently Asked Questions

How do I choose the right substitution?

Look for a function and its derivative together. Common patterns: f(g(x))·g'(x), composite functions, expressions under radicals.

What if du doesn't match exactly?

Often you can adjust by constants. If you need 2x dx but have x dx, multiply inside and divide outside by 2.

When should I use trig substitution?

For √(a²-x²) use x = a sin θ. For √(a²+x²) use x = a tan θ. For √(x²-a²) use x = a sec θ.

What's the most common mistake?

Forgetting to change dx to du, or not substituting back to x at the end.

Can I use substitution multiple times?

Yes! Some integrals require successive substitutions. Just keep track of all variables.

How do I know substitution is the right method?

If you see a composite function f(g(x)) and g'(x) (or a constant multiple) is present, try substitution.

5. Common Mistakes

Forgetting to change dx to du

Always replace dx with du using du = g'(x) dx.

Not substituting back

The final answer must be in terms of x, not u.

Wrong choice of u

Usually u should be the “inner” function whose derivative appears.

6. More Substitution Examples

Example 5.43: Exponential Composite

Problem: x2ex3dx\int x^2 e^{x^3}\,dx

Let u = x³, du = 3x² dx, so x² dx = du/3:

x2ex3dx=13eudu=13ex3+C\int x^2 e^{x^3}\,dx = \frac{1}{3}\int e^u\,du = \frac{1}{3}e^{x^3} + C
Example 5.44: Logarithm with Power

Problem: (lnx)3xdx\int \frac{(\ln x)^3}{x}\,dx

Let u = ln x, du = dx/x:

(lnx)31xdx=u3du=u44+C=(lnx)44+C\int (\ln x)^3 \cdot \frac{1}{x}\,dx = \int u^3\,du = \frac{u^4}{4} + C = \frac{(\ln x)^4}{4} + C
Example 5.45: Trigonometric Composite

Problem: sin4xcosxdx\int \sin^4 x \cos x\,dx

Let u = sin x, du = cos x dx:

u4du=u55+C=sin5x5+C\int u^4\,du = \frac{u^5}{5} + C = \frac{\sin^5 x}{5} + C
Example 5.46: Rational with Sqrt

Problem: xx2+1dx\int \frac{x}{\sqrt{x^2+1}}\,dx

Let u = x²+1, du = 2x dx:

xx2+1dx=12u1/2du=x2+1+C\int \frac{x}{\sqrt{x^2+1}}\,dx = \frac{1}{2}\int u^{-1/2}\,du = \sqrt{x^2+1} + C
Example 5.47: Secant Times Tangent

Problem: sec4xtanxdx\int \sec^4 x \tan x\,dx

Let u = sec x, du = sec x tan x dx:

sec3xsecxtanxdx=u3du=sec4x4+C\int \sec^3 x \cdot \sec x \tan x\,dx = \int u^3\,du = \frac{\sec^4 x}{4} + C
Example 5.48: Arctangent Composite

Problem: arctanx1+x2dx\int \frac{\arctan x}{1+x^2}\,dx

Let u = arctan x, du = dx/(1+x²):

udu=u22+C=(arctanx)22+C\int u\,du = \frac{u^2}{2} + C = \frac{(\arctan x)^2}{2} + C

7. Trigonometric Substitution Details

Example 5.49: √(a²-x²) Detailed

Problem: 9x2dx\int \sqrt{9-x^2}\,dx

Let x = 3 sin θ, dx = 3 cos θ dθ:

9x2=99sin2θ=3cosθ\sqrt{9-x^2} = \sqrt{9-9\sin^2\theta} = 3\cos\theta

3cosθ3cosθdθ=9cos2θdθ=92(θ+sinθcosθ)+C\int 3\cos\theta \cdot 3\cos\theta\,d\theta = 9\int \cos^2\theta\,d\theta = \frac{9}{2}(\theta + \sin\theta\cos\theta) + C

Back-substitute: θ=arcsin(x/3)\theta = \arcsin(x/3), sinθ=x/3\sin\theta = x/3, cosθ=9x2/3\cos\theta = \sqrt{9-x^2}/3

=92arcsinx3+x9x22+C= \frac{9}{2}\arcsin\frac{x}{3} + \frac{x\sqrt{9-x^2}}{2} + C
Example 5.50: √(x²+a²) Detailed

Problem: x2x2+4dx\int \frac{x^2}{\sqrt{x^2+4}}\,dx

Let x = 2 tan θ, dx = 2 sec²θ dθ:

x2+4=2secθ\sqrt{x^2+4} = 2\sec\theta, x2=4tan2θx^2 = 4\tan^2\theta

4tan2θ2sec2θ2secθdθ=4tan2θsecθdθ\int \frac{4\tan^2\theta \cdot 2\sec^2\theta}{2\sec\theta}\,d\theta = 4\int \tan^2\theta \sec\theta\,d\theta

Use tan²θ = sec²θ - 1:

=4(sec3θsecθ)dθ= 4\int (\sec^3\theta - \sec\theta)\,d\theta
Example 5.51: √(x²-a²) Detailed

Problem: x29xdx\int \frac{\sqrt{x^2-9}}{x}\,dx (x > 3)

Let x = 3 sec θ, dx = 3 sec θ tan θ dθ:

x29=3tanθ\sqrt{x^2-9} = 3\tan\theta

3tanθ3secθtanθ3secθdθ=3tan2θdθ=3(tanθθ)+C\int \frac{3\tan\theta \cdot 3\sec\theta\tan\theta}{3\sec\theta}\,d\theta = 3\int \tan^2\theta\,d\theta = 3(\tan\theta - \theta) + C

Back: =x293arcsecx3+C= \sqrt{x^2-9} - 3\text{arcsec}\frac{x}{3} + C

8. Practice Problems

1. x(x2+3)10dx\int x(x^2+3)^{10}\,dx

Answer

(x2+3)1122+C\frac{(x^2+3)^{11}}{22} + C

2. ex/xdx\int e^{\sqrt{x}}/\sqrt{x}\,dx

Answer

Let u = √x: 2ex+C2e^{\sqrt{x}} + C

3. cos(lnx)xdx\int \frac{\cos(\ln x)}{x}\,dx

Answer

sin(lnx)+C\sin(\ln x) + C

4. ex1+e2xdx\int \frac{e^x}{1+e^{2x}}\,dx

Answer

Let u = eˣ: arctan(ex)+C\arctan(e^x) + C

5. x31x4dx\int \frac{x^3}{\sqrt{1-x^4}}\,dx

Answer

Let u = x⁴: 121x4+C-\frac{1}{2}\sqrt{1-x^4} + C

6. tan5xsec2xdx\int \tan^5 x \sec^2 x\,dx

Answer

Let u = tan x: tan6x6+C\frac{\tan^6 x}{6} + C

9. Advanced Examples

Example 5.52: Double Substitution

Problem: 1xlnxdx\int \frac{1}{x\sqrt{\ln x}}\,dx

Let u = ln x, then du = dx/x. Let v = √u, then dv = du/(2√u):

1udu=2u+C=2lnx+C\int \frac{1}{\sqrt{u}}\,du = 2\sqrt{u} + C = 2\sqrt{\ln x} + C
Example 5.53: Completing the Square + Substitution

Problem: 1x2+4x+8dx\int \frac{1}{\sqrt{x^2+4x+8}}\,dx

Complete: x²+4x+8 = (x+2)² + 4. Let u = x+2:

1u2+4du=lnu+u2+4+C=lnx+2+x2+4x+8+C\int \frac{1}{\sqrt{u^2+4}}\,du = \ln|u + \sqrt{u^2+4}| + C = \ln|x+2 + \sqrt{x^2+4x+8}| + C
Example 5.54: Rationalizing Substitution

Problem: 11+xdx\int \frac{1}{1+\sqrt{x}}\,dx

Let u = √x, so x = u², dx = 2u du:

2u1+udu=(221+u)du=2u2ln1+u+C\int \frac{2u}{1+u}\,du = \int \left(2 - \frac{2}{1+u}\right)\,du = 2u - 2\ln|1+u| + C

=2x2ln(1+x)+C= 2\sqrt{x} - 2\ln(1+\sqrt{x}) + C

Example 5.55: Hyperbolic Substitution

Problem: x2+1dx\int \sqrt{x^2+1}\,dx

Let x = sinh t, dx = cosh t dt:

x2+1=cosht\sqrt{x^2+1} = \cosh t

cosh2tdt=t2+sinhtcosht2+C=sinh1x2+xx2+12+C\int \cosh^2 t\,dt = \frac{t}{2} + \frac{\sinh t \cosh t}{2} + C = \frac{\sinh^{-1}x}{2} + \frac{x\sqrt{x^2+1}}{2} + C

10. Study Tips for Substitution

1. Look for f(g(x))·g'(x)

The derivative of the inner function should appear (perhaps with a constant).

2. Common u choices

Inner function, exponent, argument of trig/log, expression under radical.

3. Adjust constants

If du has extra constants, factor them out or multiply/divide as needed.

4. Trig sub for radicals

√(a²-x²) → x=a sin θ, √(a²+x²) → x=a tan θ, √(x²-a²) → x=a sec θ

5. Complete the square

For quadratics in denominator, complete square then substitute.

6. Always back-substitute

Final answer must be in terms of original variable.

11. Substitution Patterns Summary

Integrand PatternSubstitutionResult Form
f(ax+b)f(ax+b)u = ax+b1aF(ax+b)\frac{1}{a}F(ax+b)
f(g(x))g(x)f(g(x))g'(x)u = g(x)F(g(x))F(g(x))
g(x)g(x)\frac{g'(x)}{g(x)}u = g(x)lng(x)\ln|g(x)|
a2x2\sqrt{a^2-x^2}x = a sin θTrig result
a2+x2\sqrt{a^2+x^2}x = a tan θTrig result
x2a2\sqrt{x^2-a^2}x = a sec θTrig result

12. Additional Practice

Example 5.56: Power of ln
(lnx)5xdx=(lnx)66+C\int \frac{(\ln x)^5}{x}\,dx = \frac{(\ln x)^6}{6} + C
Example 5.57: Exponential Chain
x3ex4dx=ex44+C\int x^3 e^{x^4}\,dx = \frac{e^{x^4}}{4} + C
Example 5.58: Trig Power
cos7xsinxdx=cos8x8+C\int \cos^7 x \sin x\,dx = -\frac{\cos^8 x}{8} + C
Example 5.59: Inverse Trig
(arcsinx)21x2dx=(arcsinx)33+C\int \frac{(\arcsin x)^2}{\sqrt{1-x^2}}\,dx = \frac{(\arcsin x)^3}{3} + C

13. Challenge Problems

Challenge 1:

x2(1+x3)2dx\int \frac{x^2}{(1+x^3)^2}\,dx

Solution

u = 1+x³: 13(1+x3)+C-\frac{1}{3(1+x^3)} + C

Challenge 2:

ex1+exdx\int e^x\sqrt{1+e^x}\,dx

Solution

u = 1+eˣ: 23(1+ex)3/2+C\frac{2}{3}(1+e^x)^{3/2} + C

Challenge 3:

sinx1+cos2xdx\int \frac{\sin x}{1+\cos^2 x}\,dx

Solution

u = cos x: arctan(cosx)+C-\arctan(\cos x) + C

Challenge 4:

1xx41dx\int \frac{1}{x\sqrt{x^4-1}}\,dx

Solution

u = x²: 12arcsec(x2)+C\frac{1}{2}\text{arcsec}(x^2) + C

14. More Worked Examples

Example 5.60: Exponential with Trig

Problem: etanxsec2xdx\int e^{\tan x} \sec^2 x\,dx

Let u = tan x, du = sec²x dx:

=eudu=etanx+C= \int e^u\,du = e^{\tan x} + C
Example 5.61: Rational Function

Problem: x21+x6dx\int \frac{x^2}{1+x^6}\,dx

Let u = x³, du = 3x² dx:

=1311+u2du=13arctan(x3)+C= \frac{1}{3}\int \frac{1}{1+u^2}\,du = \frac{1}{3}\arctan(x^3) + C
Example 5.62: Nested Functions

Problem: sin(x)xdx\int \frac{\sin(\sqrt{x})}{\sqrt{x}}\,dx

Let u = √x, du = dx/(2√x):

=2sinudu=2cos(x)+C= 2\int \sin u\,du = -2\cos(\sqrt{x}) + C
Example 5.63: Product Form

Problem: x(x21)99dx\int x(x^2-1)^{99}\,dx

Let u = x²-1, du = 2x dx:

=12u99du=(x21)100200+C= \frac{1}{2}\int u^{99}\,du = \frac{(x^2-1)^{100}}{200} + C
Example 5.64: Inverse Trig Composite

Problem: earctanx1+x2dx\int \frac{e^{\arctan x}}{1+x^2}\,dx

Let u = arctan x, du = dx/(1+x²):

=eudu=earctanx+C= \int e^u\,du = e^{\arctan x} + C

15. Final Practice Set

P1: 2xx2+5dx\int \frac{2x}{\sqrt{x^2+5}}\,dx

Answer

2x2+5+C2\sqrt{x^2+5} + C

P2: sin6xcosxdx\int \sin^6 x \cos x\,dx

Answer

sin7x7+C\frac{\sin^7 x}{7} + C

P3: e2x1+e4xdx\int \frac{e^{2x}}{1+e^{4x}}\,dx

Answer

12arctan(e2x)+C\frac{1}{2}\arctan(e^{2x}) + C

P4: (lnx)4xdx\int \frac{(\ln x)^4}{x}\,dx

Answer

(lnx)55+C\frac{(\ln x)^5}{5} + C

P5: sec5xtanxdx\int \sec^5 x \tan x\,dx

Answer

sec5x5+C\frac{\sec^5 x}{5} + C

P6: x4x2dx\int x\sqrt{4-x^2}\,dx

Answer

13(4x2)3/2+C-\frac{1}{3}(4-x^2)^{3/2} + C

16. Common Substitution Summary

When You SeeTry This uBecause du =
f(xn)xn1f(x^n)\cdot x^{n-1}u = x^nnx^{n-1}dx
f(ex)exf(e^x)\cdot e^xu = e^xe^x dx
f(lnx)/xf(\ln x)/xu = ln xdx/x
f(sinx)cosxf(\sin x)\cos xu = sin xcos x dx
f(cosx)sinxf(\cos x)\sin xu = cos x-sin x dx
f(tanx)sec2xf(\tan x)\sec^2 xu = tan xsec² x dx
f(secx)secxtanxf(\sec x)\sec x \tan xu = sec xsec x tan x dx
f(arctanx)/(1+x2)f(\arctan x)/(1+x^2)u = arctan xdx/(1+x²)

17. Additional Problems

x1x4dx=12arcsin(x2)+C\int \frac{x}{\sqrt{1-x^4}}\,dx = \frac{1}{2}\arcsin(x^2) + C
cosxsin2x+1dx=arctan(sinx)+C\int \frac{\cos x}{\sin^2 x + 1}\,dx = \arctan(\sin x) + C
x3x2+1dx=(x2+1)3/23(x225)+C\int x^3\sqrt{x^2+1}\,dx = \frac{(x^2+1)^{3/2}}{3}\left(x^2-\frac{2}{5}\right) + C
ex1e2xdx=arcsin(ex)+C\int \frac{e^x}{\sqrt{1-e^{2x}}}\,dx = \arcsin(e^x) + C

18. More Trig Substitution Examples

Example 5.65: ∫√(a²-x²) Type

Problem: x24x2dx\int \frac{x^2}{\sqrt{4-x^2}}\,dx

Let x = 2 sin θ:

=4sin2θdθ=2(θsinθcosθ)+C= 4\int \sin^2\theta\,d\theta = 2(\theta - \sin\theta\cos\theta) + C

=2arcsinx2x4x22+C= 2\arcsin\frac{x}{2} - \frac{x\sqrt{4-x^2}}{2} + C

Example 5.66: ∫1/√(x²+a²) Type

Problem: dxx2x2+9\int \frac{dx}{x^2\sqrt{x^2+9}}

Let x = 3 tan θ:

=19secθtan2θdθ=x2+99x+C= \frac{1}{9}\int \frac{\sec\theta}{\tan^2\theta}\,d\theta = -\frac{\sqrt{x^2+9}}{9x} + C
Example 5.67: ∫√(x²-a²) Type

Problem: x216dx\int \sqrt{x^2-16}\,dx (x > 4)

Let x = 4 sec θ:

=8tan2θsecθdθ= 8\int \tan^2\theta\sec\theta\,d\theta

Result: xx21628lnx+x216+C\frac{x\sqrt{x^2-16}}{2} - 8\ln|x + \sqrt{x^2-16}| + C

19. Completing the Square with Substitution

Example 5.68: Quadratic in Denominator

Problem: 1x26x+13dx\int \frac{1}{\sqrt{x^2-6x+13}}\,dx

Complete: x²-6x+13 = (x-3)² + 4

Let u = x-3:

=1u2+4du=lnu+u2+4+C= \int \frac{1}{\sqrt{u^2+4}}\,du = \ln|u + \sqrt{u^2+4}| + C

=lnx3+x26x+13+C= \ln|x-3 + \sqrt{x^2-6x+13}| + C

Example 5.69: √(ax²+bx+c) Form

Problem: 1x2+4x+5dx\int \frac{1}{\sqrt{-x^2+4x+5}}\,dx

Complete: -x²+4x+5 = 9-(x-2)²

=arcsinx23+C= \arcsin\frac{x-2}{3} + C

20. Chapter Summary

Basic Substitution

For ∫f(g(x))g'(x)dx, let u = g(x). The integral becomes ∫f(u)du.

Trigonometric Substitution

For expressions with √(a²-x²), √(a²+x²), or √(x²-a²), use appropriate trig substitution.

Completing the Square

For quadratic expressions, complete the square first, then substitute.

Verification

Always differentiate your answer to verify it's correct.

21. Quick Reference Card

Power Functions

  • f(xn)xn1u=xnf(x^n)x^{n-1} \to u=x^n
  • f(x)/xu=xf(\sqrt{x})/\sqrt{x} \to u=\sqrt{x}

Exponential/Log

  • f(ex)exu=exf(e^x)e^x \to u=e^x
  • f(lnx)/xu=lnxf(\ln x)/x \to u=\ln x

Trig

  • f(sinx)cosxu=sinxf(\sin x)\cos x \to u=\sin x
  • f(tanx)sec2xu=tanxf(\tan x)\sec^2 x \to u=\tan x

22. More Practice Problems

P7: x31+x2dx\int \frac{x^3}{\sqrt{1+x^2}}\,dx

Answer

(1+x2)3/231+x2+C\frac{(1+x^2)^{3/2}}{3} - \sqrt{1+x^2} + C

P8: arctanx1+x2dx\int \frac{\arctan x}{1+x^2}\,dx

Answer

(arctanx)22+C\frac{(\arctan x)^2}{2} + C

P9: sin(lnx)xdx\int \frac{\sin(\ln x)}{x}\,dx

Answer

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

P10: x21x3dx\int x^2\sqrt{1-x^3}\,dx

Answer

2(1x3)3/29+C-\frac{2(1-x^3)^{3/2}}{9} + C

23. Challenge Problems

C5: 1x+x3dx\int \frac{1}{\sqrt{x}+\sqrt[3]{x}}\,dx

Hint

Let u = x^(1/6), then x = u⁶

C6: 1+x1xdx\int \sqrt{\frac{1+x}{1-x}}\,dx

Hint

Try x = cos θ or rationalize

C7: xx4+x2+1dx\int \frac{x}{\sqrt{x^4+x^2+1}}\,dx

Hint

u = x² gives du = 2x dx

24. Detailed Worked Examples

Example 5.70: Rational with Square Root

Problem: x1+xdx\int \frac{\sqrt{x}}{1+x}\,dx

Let u = √x, so x = u², dx = 2u du:

=2u21+u2du=2(111+u2)du= 2\int \frac{u^2}{1+u^2}\,du = 2\int \left(1 - \frac{1}{1+u^2}\right)du
=2u2arctanu+C=2x2arctanx+C= 2u - 2\arctan u + C = 2\sqrt{x} - 2\arctan\sqrt{x} + C
Example 5.71: Multiple Substitutions

Problem: exxdx\int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx

Let u = √x, du = dx/(2√x):

=2eudu=2ex+C= 2\int e^u\,du = 2e^{\sqrt{x}} + C

25. Essential Substitution Results

f(ax+b)dx=1aF(ax+b)+C\int f(ax+b)\,dx = \frac{1}{a}F(ax+b) + C
f(x)f(x)dx=lnf(x)+C\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C
f(x)f(x)ndx=f(x)n+1n+1+C\int f'(x)\cdot f(x)^n\,dx = \frac{f(x)^{n+1}}{n+1} + C
f(x)ef(x)dx=ef(x)+C\int f'(x)e^{f(x)}\,dx = e^{f(x)} + C

26. Final Worked Examples

Example 5.72: Hyperbolic Substitution

Problem: 1x2+1dx\int \frac{1}{\sqrt{x^2+1}}\,dx

Let x = sinh t, dx = cosh t dt:

=coshtcoshtdt=t+C=arcsinh x+C=ln(x+x2+1)+C= \int \frac{\cosh t}{\cosh t}\,dt = t + C = \text{arcsinh } x + C = \ln(x + \sqrt{x^2+1}) + C
Example 5.73: Euler Substitution

Problem: 1x+x2+1dx\int \frac{1}{x + \sqrt{x^2+1}}\,dx

Let √(x²+1) - x = t, then x = (1-t²)/(2t):

=x2+1x+C= \sqrt{x^2+1} - x + C

27. Final Practice Problems

x41x2dx\int \frac{x^4}{\sqrt{1-x^2}}\,dx

Hint

x = sin θ

e2xex+1dx\int \frac{e^{2x}}{\sqrt{e^x+1}}\,dx

Hint

u = eˣ+1

cos(lnx)xdx\int \frac{\cos(\ln x)}{x}\,dx

Answer

sin(lnx)+C\sin(\ln x) + C

xx41dx\int \frac{x}{\sqrt{x^4-1}}\,dx

Hint

u = x² gives du = 2x dx

28. Substitution Method Selection

ExpressionSubstitution
√(a²-x²)x = a sin θ
√(a²+x²)x = a tan θ or x = a sinh t
√(x²-a²)x = a sec θ or x = a cosh t
f(g(x))·g'(x)u = g(x)
√(ax²+bx+c)Complete square first

29. Chapter Summary

Basic u-Sub

Identify inner function u and check if du is present

Trig Sub

For √(a²±x²), √(x²-a²) expressions

Complete Square

For ax²+bx+c before substitution

30. Important Standard Results

1a2x2dx=arcsinxa+C\int \frac{1}{\sqrt{a^2-x^2}}\,dx = \arcsin\frac{x}{a} + C
1x2+a2dx=lnx+x2+a2+C\int \frac{1}{\sqrt{x^2+a^2}}\,dx = \ln|x+\sqrt{x^2+a^2}| + C
1x2a2dx=lnx+x2a2+C\int \frac{1}{\sqrt{x^2-a^2}}\,dx = \ln|x+\sqrt{x^2-a^2}| + C
a2x2dx=xa2x22+a22arcsinxa+C\int \sqrt{a^2-x^2}\,dx = \frac{x\sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C

31. Final Examples

x2+a2dx=xx2+a22+a22lnx+x2+a2+C\int \sqrt{x^2+a^2}\,dx = \frac{x\sqrt{x^2+a^2}}{2} + \frac{a^2}{2}\ln|x+\sqrt{x^2+a^2}| + C
x2a2dx=xx2a22a22lnx+x2a2+C\int \sqrt{x^2-a^2}\,dx = \frac{x\sqrt{x^2-a^2}}{2} - \frac{a^2}{2}\ln|x+\sqrt{x^2-a^2}| + C
Key Takeaways

Chain Rule in Reverse

f(g(x))g(x)dx=F(g(x))+C\int f(g(x))g'(x)\,dx = F(g(x)) + C

Look for Patterns

Inner function + its derivative (or constant multiple)

Trig Substitution

For √(a²-x²), √(a²+x²), √(x²-a²)

Always Verify

Differentiate your answer to check

  • Choose u to be the “complicated” inner function
  • Don't forget to change dx to du
  • Always substitute back to original variable
  • Practice recognizing common patterns
  • Use trig substitution for radical expressions