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8.3 Trig Integrals (part 2)

Example

\[ \int \tan^3 x \, dx \]

Two ways to do subs for this example

Key Derivatives & Identities:

\[ \frac{d}{dx} \tan x = \sec^2 x \]\[ \frac{d}{dx} \sec x = \sec x \tan x \]\[ \tan^2 x + 1 = \sec^2 x \]

Just like w/ \(\cos x\) and \(\sin x\), we want to somehow bring in something to make substitution doable.

First Way:

\[ \begin{aligned} \int \tan x \cdot \tan^2 x \, dx &= \int \tan x \cdot (\sec^2 x - 1) \, dx \\ &= \int \tan x \cdot \sec^2 x \, dx - \int \tan x \, dx \end{aligned} \]

For the first integral:

\( u = \tan x \)

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

For the second integral:

\( \int \frac{\sin x}{\cos x} \, dx \)

\( U = \cos x \)

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

\[ \dots = \frac{1}{2} \tan^2 x + \ln |\cos x| + C \]

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The Second Way:

\[ \int \tan^3 x \, dx \]

Identities & Derivatives:

\[ \frac{d}{dx} \tan x = \sec^2 x \quad \frac{d}{dx} \sec x = \sec x \tan x \]\[ \tan^2 x + 1 = \sec^2 x \]

\[ \begin{aligned} &= \int \frac{\tan^3 x}{\sec x} \cdot \sec x \, dx = \int \frac{\tan^2 x}{\sec x} \cdot \tan x \sec x \, dx \\ &= \int \frac{\sec^2 x - 1}{\sec x} \cdot \sec x \tan x \, dx \end{aligned} \]

\( u = \sec x \)

\( du = \sec x \tan x \, dx \)

\[ \begin{aligned} &= \int \frac{u^2 - 1}{u} \, du = \int (u - \frac{1}{u}) \, du = \frac{1}{2} u^2 - \ln |u| + C \\ &= \frac{1}{2} \sec^2 x - \ln |\sec x| + C \end{aligned} \]

Comparison

They look different:

\[ \begin{aligned} &\frac{1}{2} (\tan^2 x + 1) - \ln |\frac{1}{\cos x}| + C \\ &= \frac{1}{2} \tan^2 x - \ln (\cos^{-1} x) + \frac{1}{2} + C \\ &= \frac{1}{2} \tan^2 x + \ln (\cos x) + B \quad \text{(where B is a constant)} \end{aligned} \]

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Example: Trigonometric Integration

\[ \int \frac{1}{\sec x - 1} dx \]

Another way to handle \(\sec x\) and \(\tan x\): turn into \(\sin x\) and \(\cos x\)

\[ = \int \frac{1}{\frac{1}{\cos x} - 1} dx = \int \frac{1}{\frac{1}{\cos x} - 1} \cdot \frac{\cos x}{\cos x} dx \]

\[ = \int \frac{\cos x}{1 - \cos x} dx = \int \frac{\cos x}{1 - \cos x} \cdot \frac{1 + \cos x}{1 + \cos x} dx \]

\[ = \int \frac{\cos x + \cos^2 x}{1 - \cos^2 x} dx = \int \frac{\cos x + \cos^2 x}{\sin^2 x} dx \]

\[ = \int \frac{\cos x}{\sin^2 x} dx + \int \frac{\cos^2 x}{\sin^2 x} dx \]

\[ = \int \frac{\cos x}{\sin^2 x} dx + \int \frac{1 - \sin^2 x}{\sin^2 x} dx = \int \frac{\cos x}{\sin^2 x} dx + \int \frac{1}{\sin^2 x} dx - \int 1 dx \]

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\[ = \int \frac{\cos x}{\sin^2 x} dx + \int \csc^2 x dx - \int 1 dx \]

Term 1 Substitution:

\( u = \sin x \)

\( du = \cos x dx \)

Term 2 Result:

\( -\cot x \)

Term 3 Result:

\( x \)

Trigonometric Derivatives

\( \frac{d}{dx} \tan x = \sec^2 x \)
\( \frac{d}{dx} \cot x = -\csc^2 x \)

Final Answer:

\[ = -\frac{1}{\sin x} - \cot x - x + C \]

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Strategy for \(\int \tan^m x \sec^n x \, dx\)

If \(n\) is even and positive

\(\rightarrow\)

save \(\sec^2 x\) and use \(u = \tan x\) along with identities

\(\sec x\) has pos. even power

If \(m\) is odd and positive

\(\rightarrow\)

save \(\sec x \tan x\) and use \(u = \sec x\) along with identities

\(\tan x\) has pos. odd power

Example

\[\int \tan x \sec^7 x \, dx\]

here, \(\tan x\) has pos. odd power (1)

save \(\sec x \tan x\)

\[= \int \sec^6 x \cdot \sec x \tan x \, dx\]

\(u = \sec x\)

\(du = \sec x \tan x \, dx\)

\[= \int u^6 \, du = \frac{1}{7} u^7 + C = \boxed{\frac{1}{7} \sec^7 x + C}\]

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Example

\(\tan x\) has pos. odd, \(\sec x\) has pos. even

\(\rightarrow\) can use either strategy

\[\int \tan^5 x \sec^6 x\]

try saving \(\sec^2 x\), \(u = \tan x\)

\[\int \tan^5 x \sec^4 x \sec^2 x \, dx\]

\(\hookrightarrow (\sec^2 x)^2 = (\tan^2 x + 1)^2\)

\[= \int \tan^5 x (\tan^2 x + 1)^2 \sec^2 x \, dx\]

\(u = \tan x\)

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

\[= \int u^5 (u^2 + 1)^2 \, du = \int u^5 (u^4 + 2u^2 + 1) \, du\]

\[= \int (u^9 + 2u^7 + u^5) \, du = \frac{1}{10} u^{10} + \frac{2}{8} u^8 + \frac{1}{6} u^6 + C\]

\[= \boxed{\frac{1}{10} \tan^{10} x + \frac{1}{4} \tan^8 x + \frac{1}{6} \tan^6 x + C}\]

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Trigonometric Integration: Alternative Substitution

Now try saving \(\sec x \tan x\) and setting \(u = \sec x\) for the integral \(\int \tan^5 x \sec^6 x \, dx\).

\[\int \tan^5 x \sec^6 x \, dx\]

We can rewrite the integral by separating one factor of \(\sec x \tan x\):

\[= \int \tan^4 x \sec^5 x \sec x \tan x \, dx\]

Using the identity \((\tan^2 x)^2 = (\sec^2 x - 1)^2\):

\[= \int (\sec^2 x - 1)^2 \sec^5 x \sec x \tan x \, dx\]

Substitution

Let \(u = \sec x\)

Then \(du = \sec x \tan x \, dx\)

Substituting \(u\) into the integral:

\[= \int (u^2 - 1)^2 u^5 \, du = \int (u^4 - 2u^2 + 1) u^5 \, du\]

\[= \int (u^9 - 2u^7 + u^5) \, du = \frac{1}{10} u^{10} - \frac{1}{4} u^8 + \frac{1}{6} u^6 + C\]

\[= \frac{1}{10} \sec^{10} x - \frac{1}{4} \sec^8 x + \frac{1}{6} \sec^6 x + C\]