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PAGE 1

8.2 Integration By Parts

NOT on exam 1

HW includes 8.1 (Basic integration)

Integration by substitution: Chain rule in reverse

Integration by parts: Product rule in reverse

\( u, v \) both functions of \( x \)

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

Integration by parts

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

in the differential form

\( d(uv) = u \, dv + v \, du \)

\( u \, dv = d(uv) - v \, du \)

integrate

\[ \int u \, dv = \int d(uv) - \int v \, du \]
\[ \int u \, dv = uv - \int v \, du \]

Given integral, identify \( u \) and \( dv \), then use \( uv - \int v \, du \)

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example

\[ \int x \cdot \ln x \, dx \]
  • not something we can integrate directly
  • substitution doesn't work either
  • now try by parts
  • identify \( u, dv \)
\[ \int x \cdot \ln x \, dx \]

two parts to consider: \( x \) and \( \ln x \)

  • u: has a simpler derivative
  • dv: easy to integrate

a good rule of thumb for order of choosing u

LIATE

  • Logarithmic
  • Inverse trig
  • Algebraic (e.g. \( x^2 \))
  • Trig
  • Exponential

pick in that order

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Integration by Parts: Example Continued

Back to the integral:

\[ \int x \cdot \ln x \, dx \]

Identifying the parts:

  • \( x \) is algebraic
  • \( \ln x \) is logarithmic

LIATE: pick \( u = \ln x \) since it's L which is before A (\( x \)).

\( u = \ln x \)

Differentiate:

\( \frac{du}{dx} = \frac{1}{x} \)

\( du = \frac{1}{x} \, dx \)

\( dv = x \, dx \) → any left over after choosing \( u \)

Integrate:

\( v = \int dv \)

\( v = \int x \, dx = \frac{1}{2}x^2 \)

don't need + C

Formula:

\[ \int u \, dv = uv - \int v \, du \]
\[ = (\ln x)(\frac{1}{2}x^2) - \int (\frac{1}{2}x^2)(\frac{1}{x} \, dx) \]\[ = \frac{1}{2}x^2 \ln x - \int \frac{1}{2}x \, dx \]
\[ = \frac{1}{2}x^2 \ln x - \frac{1}{4}x^2 + C \]

to check: take derivative, it must match \( x \ln x \)

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Example: Trigonometric Integration by Parts

\[ \text{example } \int x \cos x \, dx \]

Parts: \( x \) and \( \cos x \)

  • \( x \): algebraic (A)
  • \( \cos x \): trig (T)

Order for \( u \): LIATE. A before T so we choose \( u = x \).

deriv.

\( u = x \)

\( du = dx \)

integrate

\( dv = \cos x \, dx \)

\( v = \sin x \)

+ C is not needed here

\[ uv - \int v \, du \]\[ = x \sin x - \int \sin x \, dx = x \sin x - (-\cos x) + C \]
\[ = x \sin x + \cos x + C \]

Definite Integral

\[ \int_{0}^{\pi/2} x \cos x \, dx \]

pick \( u, dv \) as usual

\[ \text{then } uv \Big|_0^{\pi/2} - \int_0^{\pi/2} v \, du = x \sin x \Big|_0^{\pi/2} - \int_0^{\pi/2} \sin x \, dx \]
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\[ = \frac{\pi}{2} \sin \frac{\pi}{2} + \cos x \Big|_0^{\pi/2} = \frac{\pi}{2} + \cos\left(\frac{\pi}{2}\right) - \cos(0) = \boxed{\frac{\pi}{2} - 1} \]

Example

\[ \int x^2 \cos x \, dx \]

LIATE

so \( u = x^2 \) (A before T)

\( u = x^2 \)

\( du = 2x \, dx \)

\( dv = \cos x \, dx \)

\( v = \sin x \)

\[ uv - \int v \, du \]\[ = x^2 \sin x - \int (\sin x)(2x) \, dx \]\[ = x^2 \sin x - 2 \int x \cdot \sin x \, dx \]

By parts again:

\( x \to A \), \( \sin x \to T \)

\( U = x \)

\( dV = \sin x \, dx \)

\( dU = dx \)

\( V = -\cos x \)

\( UV - \int V \, dU \)

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\[ = x^2 \sin x - 2 \left( UV - \int V \, dU \right) \]\[ = x^2 \sin x - 2 \left( -x \cos x - \int -\cos x \, dx \right) \]\[ = x^2 \sin x - 2 \left( -x \cos x + \int \cos x \, dx \right) \]\[ = x^2 \sin x - 2 (-x \cos x + \sin x) + C \]\[ = \boxed{x^2 \sin x + 2x \cos x - 2 \sin x + C} \]

General Rule:

\( \int x^n \cos x \, dx \to n \) rounds

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Example: Integration by Parts with Cyclic Integrals

Consider the integral:

\[ \int e^x \cos x \, dx \]

LIATE Rule Application:

In the expression \( e^x \cos x \), \( e^x \) is Exponential (E) and \( \cos x \) is Trigonometric (T). According to LIATE, T comes before E, so we choose:

\( u = \cos x \)

\( u = \cos x \)

\( du = -\sin x \, dx \)

\( dv = e^x \, dx \)

\( v = e^x \)

Applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \):

\[ \begin{aligned} \int e^x \cos x \, dx &= e^x \cos x - \int e^x (-\sin x) \, dx \\ &= e^x \cos x + \int e^x \sin x \, dx \end{aligned} \]

Second Application of Integration by Parts

We apply integration by parts again to the new integral \( \int e^x \sin x \, dx \):

\( U = \sin x \)

\( dU = \cos x \, dx \)

\( dV = e^x \, dx \)

\( V = e^x \)

\[ \begin{aligned} \int e^x \cos x \, dx &= e^x \cos x + \left( UV - \int V \, dU \right) \\ &= e^x \cos x + e^x \sin x - \int e^x \cos x \, dx \end{aligned} \]

Observation:

The original integral has reappeared. Continuing integration by parts would result in an infinite loop. STOP!

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Solving the Algebraic Equation

We now have an equation where the integral appears on both sides:

\[ \int e^x \cos x \, dx = e^x \cos x + e^x \sin x - \int e^x \cos x \, dx \]

Add \( \int e^x \cos x \, dx \) to both sides:

\[ 2 \int e^x \cos x \, dx = e^x (\cos x + \sin x) \]

So, dividing by 2 and adding the constant of integration \( C \):

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

Alternative Approach

This is an example of a case where going against the LIATE rule still works:

\( u = e^x \)

\( du = e^x \, dx \)

\( dv = \cos x \, dx \)

\( v = \sin x \)

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Exam 1 Review Session

SEPT 18th

(Monday)

5:30pm-7:30pm

MATH 175

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