MathIsimple
Course 10

Integration Techniques

Section 1: Integration by Substitution

Theorem 1.1: Substitution Method

If u=g(x)u = g(x) is differentiable and FF is an antiderivative of ff, then:

∫f(g(x))g′(x) dx=∫f(u) du=F(u)+C=F(g(x))+C\int f(g(x))g'(x)\,dx = \int f(u)\,du = F(u) + C = F(g(x)) + C
Example 1.1: Basic Substitution

Find: ∫2x(x2+1)5 dx\int 2x(x^2+1)^5\,dx

Solution: Let u=x2+1u = x^2 + 1, then du=2x dxdu = 2x\,dx.

∫2x(x2+1)5 dx=∫u5 du=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 1.2: Trigonometric Substitution

Find: ∫cos⁡x⋅esin⁡x dx\int \cos x \cdot e^{\sin x}\,dx

Solution: Let u=sin⁡xu = \sin x, then du=cos⁡x dxdu = \cos x\,dx.

∫cos⁡x⋅esin⁡x dx=∫eu du=eu+C=esin⁡x+C\int \cos x \cdot e^{\sin x}\,dx = \int e^u\,du = e^u + C = e^{\sin x} + C

Section 2: Integration by Parts

Theorem 2.1: Integration by Parts

If uu and vv are differentiable functions, then:

∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du
Proof of Integration by Parts:

From the product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'

Integrating both sides:

uv=∫u′v dx+∫uv′ dxuv = \int u'v\,dx + \int uv'\,dx

Rearranging:

∫uv′ dx=uv−∫u′v dx\int uv'\,dx = uv - \int u'v\,dx

With dv=v′ dxdv = v'\,dx and du=u′ dxdu = u'\,dx, we get the formula. ∎

∎
Example 2.1: Integration by Parts

Find: ∫xex dx\int x e^x\,dx

Solution: Let u=xu = x, dv=ex dxdv = e^x\,dx.

Then du=dxdu = dx, v=exv = e^x.

∫xex dx=xex−∫ex dx=xex−ex+C=ex(x−1)+C\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1) + C
Example 2.2: Logarithmic Function

Find: ∫ln⁡x dx\int \ln x\,dx

Solution: Let u=ln⁡xu = \ln x, dv=dxdv = dx.

Then du=dxxdu = \frac{dx}{x}, v=xv = x.

∫ln⁡x dx=xln⁡x−∫x⋅1x dx=xln⁡x−x+C\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - x + C

Section 3: Special Techniques

Example 3.1: Repeated Integration by Parts

Find: ∫x2ex dx\int x^2 e^x\,dx

Solution: Apply parts twice:

First: u=x2u = x^2, dv=ex dxdv = e^x\,dx → ∫x2ex dx=x2ex−2∫xex dx\int x^2 e^x\,dx = x^2 e^x - 2\int x e^x\,dx

Second: u=xu = x, dv=ex dxdv = e^x\,dx → ∫xex dx=xex−ex\int x e^x\,dx = xe^x - e^x

∫x2ex dx=x2ex−2xex+2ex+C=ex(x2−2x+2)+C\int x^2 e^x\,dx = x^2 e^x - 2xe^x + 2e^x + C = e^x(x^2 - 2x + 2) + C
Example 3.2: Cycling Integration by Parts

Find: ∫exsin⁡x dx\int e^x \sin x\,dx

Solution: Let I=∫exsin⁡x dxI = \int e^x \sin x\,dx.

Apply parts twice, then solve for II:

I=exsin⁡x−excos⁡x−II = e^x \sin x - e^x \cos x - I
2I=ex(sin⁡x−cos⁡x)2I = e^x(\sin x - \cos x)
I=ex(sin⁡x−cos⁡x)2+CI = \frac{e^x(\sin x - \cos x)}{2} + C

Section 4: Trigonometric Substitution

Example 4.1: Trigonometric Substitution

Find: ∫11−x2 dx\int \frac{1}{\sqrt{1-x^2}}\,dx

Let x=sin⁡θx = \sin \theta, then dx=cos⁡θ dθdx = \cos \theta\,d\theta:

∫cos⁡θcos⁡θ dθ=θ+C=arcsin⁡x+C\int \frac{\cos \theta}{\cos \theta}\,d\theta = \theta + C = \arcsin x + C

Section 5: Partial Fractions

Example 5.1: Partial Fractions

Find: ∫1x2−1 dx\int \frac{1}{x^2-1}\,dx

1x2−1=12(1x−1−1x+1)\frac{1}{x^2-1} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right)

∫=12(ln⁡∣x−1∣−ln⁡∣x+1∣)+C=12ln⁡∣x−1x+1∣+C\int = \frac{1}{2}(\ln|x-1| - \ln|x+1|) + C = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C

Section 6: Rationalizing Substitutions

Example 6.1: Rationalizing Substitution

Find: ∫11+x dx\int \frac{1}{1+\sqrt{x}}\,dx

Let u=xu = \sqrt{x}, then x=u2x = u^2, dx=2u dudx = 2u\,du:

∫2u1+u du=2∫(1−11+u)du=2(u−ln⁡∣1+u∣)+C\int \frac{2u}{1+u}\,du = 2\int \left(1 - \frac{1}{1+u}\right)du = 2(u - \ln|1+u|) + C

Section 7: Reduction Formulas

Example 7.1: Reduction Formula

For In=∫sin⁡nx dxI_n = \int \sin^n x\,dx:

In=−sin⁡n−1xcos⁡xn+n−1nIn−2I_n = -\frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} I_{n-2}
Practice Quiz: Integration Techniques
10
Questions
0
Correct
0%
Accuracy
1
∫2x(x2+1)5 dx\int 2x(x^2+1)^5\,dx using u=x2+1u = x^2+1:
Easy
Not attempted
2
∫xex dx=\int x e^x\,dx =
Easy
Not attempted
3
∫ln⁡x dx=\int \ln x\,dx =
Easy
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4
∫cos⁡x⋅esin⁡x dx=\int \cos x \cdot e^{\sin x}\,dx =
Easy
Not attempted
5
∫x2ex dx\int x^2 e^x\,dx requires how many applications?
Medium
Not attempted
6
∫tan⁡x dx=∫sin⁡xcos⁡x dx=\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx =
Medium
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7
∫xcos⁡x dx=\int x \cos x\,dx =
Medium
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8
For ∫a2−x2 dx\int \sqrt{a^2-x^2}\,dx, what substitution?
Hard
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9
∫exsin⁡x dx\int e^x \sin x\,dx requires:
Hard
Not attempted
10
∫ln⁡xx dx=\int \frac{\ln x}{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 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 should I use substitution vs integration by parts?

Use substitution when you see a composite function f(g(x)) with g'(x) present. Use parts for products of different function types (e.g., x·e^x, x·sin x).

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.