Download original notes
PAGE 1

8.4 Trigonometric Substitution (part 1)

\[ \int \frac{1}{\sqrt{1-x^2}} dx \]

trig substitution: \( x = \text{some trig function} \)

let \( x = \sin \theta \) → why? we'll see in the following example

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

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

now back to x

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

\[ \sin^{-1}(x) + C \]
PAGE 2

example

\[ \int \frac{x^3}{\sqrt{x^2+4}} dx \]

to determine the right trig subs, look at the radical part

\[ \sqrt{x^2+4} \]

set up a triangle with sides \( \sqrt{x^2+4}, x, 2 \)

  • \( \sqrt{x^2+4} \) is a square root
  • \( x \) is a square root
  • \( 2 \) is a square root
Right triangle with angle \theta, opposite side x, adjacent side 2, and hypotenuse \sqrt{x^2+4}.

hypotenuse: \( \sqrt{x^2+4} \) because

\[ (\sqrt{x^2+4})^2 = x^2 + 2^2 \]

other two sides

two remaining sides: \( x, 2 \)

rule of thumb: place constant as adjacent

PAGE 3

Relate \( x \) and \( \theta \) in the simplest way using the triangle.

Use the simpler sides

Right triangle with angle theta, opposite side x, adjacent side 2, and hypotenuse sqrt(x^2+4).

Simplest way: \( \tan \theta = \frac{x}{2} \)

\( x = 2 \tan \theta \)

\( dx = 2 \sec^2 \theta \, d\theta \)

\[ \int \frac{x^3}{\sqrt{x^2+4}} \, dx = \int \frac{(2 \tan \theta)^3}{\sqrt{(2 \tan \theta)^2 + 4}} \cdot 2 \sec^2 \theta \, d\theta \]
\[ = \int \frac{8 \tan^3 \theta}{\sqrt{4 \tan^2 \theta + 4}} \cdot 2 \sec^2 \theta \, d\theta \]

\( \sqrt{4(\tan^2 \theta + 1)} = 2 \sqrt{\sec^2 \theta} = 2 \sec \theta \)

\( \tan^2 \theta + 1 = \sec^2 \theta \)

\[ = \int \frac{8 \tan^3 \theta}{2 \sec \theta} \cdot 2 \sec^2 \theta \, d\theta = 8 \int \tan^3 \theta \sec \theta \, d\theta \]
PAGE 4
\[ = 8 \int \tan^2 \theta \sec \theta \tan \theta \, d\theta \]

\( u = \sec \theta \)

\( du = \sec \theta \tan \theta \, d\theta \)

\[ = 8 \int (\sec^2 \theta - 1) \sec \theta \tan \theta \, d\theta \]
\[ = 8 \int (u^2 - 1) \, du = 8 \left( \frac{u^3}{3} - u \right) + C = \frac{8}{3} \sec^3 \theta - 8 \sec \theta + C \]

Not done yet, go back to \( x \)

Bring triangle back:

Right triangle with angle theta, opposite side x, adjacent side 2, and hypotenuse sqrt(x^2+4).

\( \sec \theta = \frac{1}{\cos \theta} = \frac{\text{hyp}}{\text{adj}} = \frac{\sqrt{x^2+4}}{2} \)

\[ = \frac{8}{3} \left( \frac{\sqrt{x^2+4}}{2} \right)^3 - 8 \left( \frac{\sqrt{x^2+4}}{2} \right) + C \]

\[ = \frac{1}{3} (x^2+4)^{3/2} - 4(x^2+4)^{1/2} + C \]

PAGE 5

Trigonometric Substitution Example

Example

\[ \int \frac{1}{(1-x^2)^{3/2}} dx = \int \frac{1}{(\sqrt{1-x^2})^3} dx \]

Construct triangle w/ sides: \(\sqrt{1-x^2}\), \(1\), \(x\)

Note: \(\sqrt{1-x^2}\) is a root, and \(1\), \(x\) are roots.

Determine Hypotenuse

\(1\), because \(1^2 = (\sqrt{1-x^2})^2 + x^2\)

Remaining sides:

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

Neither is constant, so we take the one containing a constant as adjacent.

Right triangle with hypotenuse 1, opposite side x, adjacent side \sqrt{1-x^2}, and angle \theta.
PAGE 6

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

\[ \sin \theta = \frac{x}{1} \]\[ x = \sin \theta \]\[ dx = \cos \theta d\theta \]
\[ \int \frac{1}{(1-x^2)^{3/2}} dx = \int \frac{1}{(1-\sin^2 \theta)^{3/2}} \cos \theta d\theta \]\[ = \int \frac{1}{(\cos^2 \theta)^{3/2}} \cos \theta d\theta = \int \frac{1}{\cos^3 \theta} \cos \theta d\theta = \int \frac{1}{\cos^2 \theta} d\theta \]\[ = \int \sec^2 \theta d\theta = \tan \theta + C \]

Go back to \(x\):

\[ = \frac{x}{\sqrt{1-x^2}} + C \]
Right triangle with hypotenuse 1, opposite side x, adjacent side \sqrt{1-x^2}, and angle \theta.
\[ \tan \theta = \frac{\text{opp}}{\text{adj}} = \frac{x}{\sqrt{1-x^2}} \]
PAGE 7

Example: Trigonometric Substitution

\[ \int_{0}^{2} \frac{x^2}{x^2+4} dx = \int_{0}^{2} \frac{x^2}{(\sqrt{x^2+4})^2} dx \]

triangle w/ sides: \(\sqrt{x^2+4}, x, 2\)

hypotenuse: \(\sqrt{x^2+4}\) adjacent: 2

Right triangle with hypotenuse \sqrt{x^2+4}, opposite side x, adjacent side 2, and angle \theta.

\(\tan \theta = \frac{x}{2}\)

\(x = 2 \tan \theta\)

\(dx = 2 \sec^2 \theta d\theta\)

\(\theta = \tan^{-1}(\frac{x}{2})\)

Change Integration Limits to Refer to \(\theta\)

  • upper limit: \(x=2 \rightarrow \theta = \tan^{-1}(\frac{2}{2}) = \tan^{-1}(1) = \frac{\pi}{4}\)
  • lower limit: \(x=0 \rightarrow \theta = \tan^{-1}(\frac{0}{2}) = \tan^{-1}(0) = 0\)
PAGE 8
\[ \int_{0}^{2} \frac{x^2}{x^2+4} dx = \int_{0}^{\pi/4} \frac{(2 \tan \theta)^2}{(2 \tan \theta)^2 + 4} \cdot 2 \sec^2 \theta d\theta \]
\[ = \int_{0}^{\pi/4} \frac{4 \tan^2 \theta}{4(\tan^2 \theta + 1)} \cdot 2 \sec^2 \theta d\theta = \int_{0}^{\pi/4} \frac{4 \tan^2 \theta}{4 \sec^2 \theta} \cdot 2 \sec^2 \theta d\theta \]
\[ = 2 \int_{0}^{\pi/4} \tan^2 \theta d\theta = 2 \int_{0}^{\pi/4} (\sec^2 \theta - 1) d\theta \]
\[ = 2 \left( \int_{0}^{\pi/4} \sec^2 \theta d\theta - \int_{0}^{\pi/4} d\theta \right) \]
\[ = 2 \left( \tan \theta \Big|_0^{\pi/4} - \theta \Big|_0^{\pi/4} \right) = 2 \left( 1 - 0 - \frac{\pi}{4} - 0 \right) = \boxed{2(1 - \frac{\pi}{4})} \]