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5.5 Substitution Rule (part 2)

General idea: choose \( u \) such that its derivative is a constant-multiple of the left over parts

Example

\[ \int \sin^3(2x) \cos(2x) \, dx \]

\[ = \int [\sin(2x)]^3 \cdot \cos(2x) \, dx \]

Note on choosing \( u \):

Ignore the exponent (3) while choosing \( u \).
Choose \( u \) from the base functions: \( \sin(2x) \) and \( \cos(2x) \).

Compare \( \sin(2x) \) and \( \cos(2x) \). Notice the derivative of \( \sin(2x) \) is \( 2 \cos(2x) \), which is two times the left over part \( (\cos(2x)) \).

So, let \( u = \sin(2x) \)

then \( \frac{du}{dx} = 2 \cos(2x) \)

\( du = 2 \cos(2x) \, dx \)

Sub these into integral, sub \( x \) and \( dx \) out

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\[ \int \underbrace{[\sin(2x)]^3}_{u^3} \cdot \underbrace{\cos(2x) \, dx}_{\frac{1}{2} du} \]

\( u = \sin(2x) \)

\( du = 2 \cos(2x) \, dx \)

I have this but without the 2:

\( \frac{1}{2} du = \cos(2x) \, dx \)

\[ = \int \frac{1}{2} u^3 \, du = \frac{1}{2} \int u^3 \, du = \frac{1}{2} \left( \frac{u^4}{4} \right) + C \]

now sub \( u \) out

\[ = \frac{1}{8} [\sin(2x)]^4 + C = \boxed{\frac{1}{8} \sin^4(2x) + C} \]

revisit the start again

\[ \int [\sin(2x)]^3 \cos(2x) \, dx \]

Compare: \( \sin(2x) \) and \( \cos(2x) \)

We chose \( u = \sin(2x) \) because its derivative \( 2 \cos(2x) \) is 2 times the other part.

But notice derivative of \( \cos(2x) \) is \( -2 \sin(2x) \) which is also a constant multiple of the other part \( (\sin(2x)) \).

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Alternative Substitution Strategy

So what if we had chosen \( u = \cos(2x) \) instead?

\[ \int [\sin(2x)]^3 \cos(2x) \, dx \]

\( u = \cos(2x) \)

\( \frac{du}{dx} = -2 \sin(2x) \)

\( du = -2 \sin(2x) \, dx \)

\[ = \int \underbrace{\cos(2x)}_{u} \cdot \underbrace{[\sin(2x)]^3}_{?} \, dx \]

What to do with this? There is no \( [\sin(2x)]^3 \) in \( du \).

Rewrite:

\[ \int \underbrace{\cos(2x)}_{u} \cdot \underbrace{[\sin(2x)]^2}_{1-u^2} \cdot \underbrace{\sin(2x) \, dx}_{-\frac{1}{2} du} \]

\( \sin^2(2x) + \cos^2(2x) = 1 \)

\( \sin^2(2x) + u^2 = 1 \)

\( \sin^2(2x) = 1 - u^2 \)

\( \sin(2x) \, dx = -\frac{1}{2} du \)

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New Integral in \( u \)

\[ \int u \cdot (1 - u^2) \cdot -\frac{1}{2} \, du \]

\[ = -\frac{1}{2} \int u(1 - u^2) \, du = -\frac{1}{2} \int (u - u^3) \, du \]

\[ = -\frac{1}{2} \left( \frac{u^2}{2} - \frac{u^4}{4} \right) + C = -\frac{1}{4} u^2 + \frac{1}{8} u^4 + C \]

\( u = \cos(2x) \)

\[ = -\frac{1}{4} \cos^2(2x) + \frac{1}{8} \cos^4(2x) + C \]

Can use trig identities to make it look like the previous answer.

If by the time you try to get rid of \( x \) and sub in \( u \) things get messy, consider changing choice of \( u \).

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Example: Integration by Substitution

\[ \int \frac{e^{2x}}{e^{2x} + 3} \, dx \]

\[ = \int (e^{2x} + 3)^{-1} (e^{2x}) \, dx \]

Note: Ignore the power for now and compare the base parts to choose \( u \).

\[ \frac{d}{dx}(e^{2x} + 3) = 2e^{2x} \] (which is 2 times the leftover \( e^{2x} \))

but \[ \frac{d}{dx}(e^{2x}) = 2e^{2x} \neq \text{constant multiple of } e^{2x} + 3 \]

So, we pick \( u = e^{2x} + 3 \)

\[ \frac{du}{dx} = 2e^{2x} \]

\[ du = 2e^{2x} \, dx \]

\[ \frac{1}{2} du = e^{2x} \, dx \]

\[ \int \underbrace{(e^{2x} + 3)^{-1}}_{u^{-1}} \underbrace{(e^{2x}) \, dx}_{\frac{1}{2} du} \]

\[ = \frac{1}{2} \int u^{-1} \, du = \frac{1}{2} \ln |u| + C = \boxed{\frac{1}{2} \ln |e^{2x} + 3| + C} \]

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Example: Trigonometric Integration

\[ \int \frac{\sec^2 x}{\tan^3 x} \, dx \]

\[ = \int (\tan x)^{-3} (\sec x)^2 \, dx \]

If we temporarily ignore the powers as we usually do, we compare \( \tan x \) and \( \sec x \):

\[ \frac{d}{dx} \tan x = \sec^2 x \neq \text{constant multiple of leftover } (\sec x) \]
\[ \frac{d}{dx} \sec x = \sec x \tan x \neq \text{constant multiple of } \tan x \]

No appropriate choice of \( u \) found this way.

\[ \int (\tan x)^{-3} (\sec^2 x) \, dx \]

Now let's keep the power of \( \sec x \):

\[ \frac{d}{dx} \tan x = \sec^2 x \] which matches the other part exactly.

So, let \( u = \tan x \)

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\(\frac{du}{dx} = \sec^2 x\)

\(du = \sec^2 x \, dx\)

\[ \int \underbrace{(\tan x)^{-3}}_{u^{-3}} \underbrace{(\sec^2 x) \, dx}_{du} = \int u^{-3} \, du \]

\[ = \frac{u^{-2}}{-2} + C = -\frac{1}{2} u^{-2} + C = -\frac{1}{2u^2} + C \]

\[ = -\frac{1}{2 \tan^2 x} + C \]

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example

\[ \int_{e^{36}}^{e^{81}} \frac{1}{x \sqrt{\ln x}} \, dx \]

\[ = \int_{e^{36}}^{e^{81}} (\ln x)^{-1/2} \left( \frac{1}{x} \right) \, dx \]

we notice \(\frac{d}{dx} \ln x = \frac{1}{x}\) so we let \(u = \ln x\)

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

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

now we adjust the integration limits to refer to \(u\)

old upper limit: \(x = e^{81} \rightarrow u = \ln x = \ln e^{81} = \) 81 new upper limit

old lower limit: \(x = e^{36} \rightarrow u = \ln x = \ln e^{36} = \) 36 new lower limit

now sub in \(u\), \(du\), and the new limits

\[ \int_{36}^{81} u^{-1/2} \, du = \left. \frac{u^{1/2}}{1/2} \right|_{36}^{81} \]

Do NOT go back to \(x\)

we changed the limits to refer to \(u\) so stay w/ \(u\)

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Definite Integral Evaluation

The following calculation shows the final steps of evaluating a definite integral, likely involving a square root substitution or power rule application.

\[ = 2u^{1/2} \Big|_{36}^{81} = 2(81)^{1/2} - 2(36)^{1/2} = 2(9) - 2(6) = 18 - 12 = 6 \]

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