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8.4 Trig Subs (part 2)

Example

Find the length of \( y = \sqrt{1-x^2} \) from \( x = 0 \) to \( x = \frac{1}{\sqrt{2}} \).

\[ L = \int_{a}^{b} \sqrt{1 + (y')^2} \, dx \]

\( y = (1-x^2)^{1/2} \)

\( y' = \frac{1}{2} (1-x^2)^{-1/2} (-2x) = \frac{-x}{\sqrt{1-x^2}} \)

\[ 1 + (y')^2 = 1 + \left( \frac{-x}{\sqrt{1-x^2}} \right)^2 = 1 + \frac{x^2}{1-x^2} = \frac{1}{1-x^2} \]
\[ L = \int_{0}^{\frac{1}{\sqrt{2}}} \sqrt{\frac{1}{1-x^2}} \, dx = \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{\sqrt{1-x^2}} \, dx \]

Now knowing trig subs we can evaluate this

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Triangle with sides \( \sqrt{1-x^2}, 1, x \)

  • Hypotenuse: \( 1 \)
  • Adjacent: \( \sqrt{1-x^2} \) (part containing a constant)

Relate \( x \) and \( \theta \) using simplest sides:

Right triangle with hypotenuse 1, opposite side x, adjacent side \sqrt{1-x^2}, and angle \theta.

\( \sin \theta = \frac{x}{1} \)

\( x = \sin \theta \rightarrow \theta = \sin^{-1}(x) \)

\( dx = \cos \theta \, d\theta \)

Upper limit: \( x = \frac{1}{\sqrt{2}} \rightarrow \theta = \sin^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4} \)

Lower limit: \( x = 0 \rightarrow \theta = \sin^{-1}(0) = 0 \)

\[ \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{\sqrt{1-x^2}} \, dx = \int_{0}^{\frac{\pi}{4}} \frac{1}{\sqrt{1-\sin^2 \theta}} \cos \theta \, d\theta \]
\[ = \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos \theta} \cos \theta \, d\theta = \int_{0}^{\frac{\pi}{4}} d\theta = \theta \Big|_{0}^{\frac{\pi}{4}} = \frac{\pi}{4} \]
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Geometrically, this was part of the circumference of a circle radius 1.

A graph of a unit circle arc in the first quadrant from x=0 to x=1/sqrt(2) with length pi/4.

length = \(\frac{\pi}{4}\)

We cannot have other hypotenuse, both adj and opp can be swapped.

Right triangle with hypotenuse 1, base x, height sqrt(1-x^2), and angle theta.

Relate \(x\) and \(\theta\):

\[ \cos \theta = \frac{x}{1} \]
  • \(x = \cos \theta \rightarrow \theta = \cos^{-1}(x)\)
  • \(dx = -\sin \theta \, d\theta\)
  • \(x = \frac{1}{\sqrt{2}} \rightarrow \theta = \cos^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}\)
  • \(x = 0 \rightarrow \theta = \cos^{-1}(0) = \frac{\pi}{2}\)
\[ \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{\sqrt{1-x^2}} dx = \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} \frac{1}{\sqrt{1-\cos^2 \theta}} \cdot -\sin \theta \, d\theta \]
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\[ = \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} \frac{1}{\sin \theta} \cdot -\sin \theta \, d\theta = \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} -d\theta = -\theta \Big|_{\frac{\pi}{2}}^{\frac{\pi}{4}} = -\frac{\pi}{4} + \frac{\pi}{2} = \frac{\pi}{4} \]

The rule of thumb of putting constant or constant-containing part as adjacent limits the subs to 3: \(\sin \theta, \sec \theta, \tan \theta\)

Swapping adj and opp brings in additional 3: \(\cos \theta, \csc \theta, \cot \theta\)

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Example: Integration by Completing the Square

\[ \int \frac{1}{x^2 - 6x + 45} \, dx = \int \frac{1}{(\sqrt{x^2 - 6x + 45})^2} \, dx \]

Notice the part under the radical doesn't look like a sum or difference of squares \(\rightarrow\) useful for building triangles.

Complete the square:

\[ \begin{aligned} x^2 - 6x + 45 &= x^2 - 6x + (\frac{-6}{2})^2 + 45 - (\frac{-6}{2})^2 \\ &= x^2 - 6x + 9 + 45 - 9 \\ &= (x - 3)^2 + 6^2 \end{aligned} \]
\[ \int \frac{1}{(\sqrt{(x-3)^2 + 6^2})^2} \, dx \]
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\[ \int \frac{1}{(\sqrt{(x-3)^2 + 6^2})^2} \, dx \]

\( u = x - 3 \)

\( du = dx \)

\[ \int \frac{1}{(\sqrt{u^2 + 6^2})^2} \, du \]

Triangle with sides \( \sqrt{u^2 + 6^2}, u, 6 \)

  • Hypotenuse: \( \sqrt{u^2 + 6^2} \)
  • Adjacent: \( 6 \)
Right triangle with angle theta, adjacent side 6, opposite side u, and hypotenuse sqrt(u^2 + 6^2).

Via \( \tan \theta = \frac{u}{6} \rightarrow u = 6 \tan \theta \)

\( du = 6 \sec^2 \theta \, d\theta \)

\[ \int \frac{1}{(\sqrt{36 \tan^2 \theta + 36})^2} \cdot 6 \sec^2 \theta \, d\theta = \int \frac{1}{36 \sec^2 \theta} \cdot 6 \sec^2 \theta \, d\theta \]

\( 36(\tan^2 \theta + 1) \)

\( 36 \sec^2 \theta \)

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\[ = \int \frac{1}{6} d\theta = \frac{1}{6} \theta + C \]

how back to \( u \), back to \( x \)

from \( u = 6 \tan \theta \quad \rightarrow \quad \theta = \tan^{-1}\left( \frac{u}{6} \right) = \tan^{-1}\left( \frac{x-3}{6} \right) \)

\[ = \frac{1}{6} \tan^{-1}\left( \frac{x-3}{6} \right) + C \]

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example

\[ \int \frac{1}{\sqrt{(x-5)(1-x)}} dx \]

make what's under radical a sum or difference of squares

\[ (x-5)(1-x) = -x^2 + 6x - 5 \]\[ = -(x^2 - 6x + 5) \quad \text{complete square inside ( )} \]\[ = -(x^2 - 6x + 9 + 5 - 9) \]\[ = -\left[ (x^2 - 6x + 9) - 4 \right] = -\left[ (x-3)^2 - 2^2 \right] \]\[ = 2^2 - (x-3)^2 \]
\[ \int \frac{1}{\sqrt{2^2 - (x-3)^2}} dx \]
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Trigonometric Substitution: Example 2

Consider a triangle with sides:

\[ \sqrt{4 - (x-3)^2}, \quad 2, \quad x-3 \]
  • hypotenuse: \( 2 \)
  • adjacent: \( \sqrt{4 - (x-3)^2} \)
Right triangle with hypotenuse 2, opposite side x-3, adjacent side sqrt(4-(x-3)^2), and angle theta.
\[ \sin \theta = \frac{x-3}{2} \]\[ x-3 = 2 \sin \theta \]\[ x = 2 \sin \theta + 3 \]\[ dx = 2 \cos \theta \, d\theta \]

Integration Process

\[ \int \frac{1}{\sqrt{4 - (x-3)^2}} \, dx = \int \frac{1}{\sqrt{4 - 4 \sin^2 \theta}} \cdot 2 \cos \theta \, d\theta = \int \frac{1}{\sqrt{4 \cos^2 \theta}} \cdot 2 \cos \theta \, d\theta \]

Note on simplification:

\[ 4(1 - \sin^2 \theta) = 4 \cos^2 \theta \]
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\[ = \int d\theta = \theta + C \]

From our initial substitution:

\[ \sin \theta = \frac{x-3}{2} \implies \theta = \sin^{-1} \left( \frac{x-3}{2} \right) \]

Final Result

\[ = \sin^{-1} \left( \frac{x-3}{2} \right) + C \]