Trigonometric Substitutions

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Rough meaning of grades:
$\qquad$ 29-34 is A
$\qquad$ 23-28 is B
$\qquad$ 17-22 is C
$\qquad$ 11-16 is D
Regarding the quiz--if you do every homework problem that was assigned, you'll have a severe case of deja vu on the quiz! On the exam, we do not restrict ourselves like this, but you get to have a sheet of paper.

The first homework problem is to compute

$$\int_{\sqrt{2}}^2 \frac{1}{x^3 \sqrt{x^2-1}} dx.$$ (5.4)

Your first idea might be to do some sort of substitution, e.g., $u=x^2-1$, but $du=2xdx$ is nowhere to be seen and this simply doesn't work. Likewise, integration by parts gets us nowhere. However, a technique called ``inverse trig substitutions'' and a trig identity easily dispenses with the above integral and several similar ones! Here's the crucial table:

Expression	Inverse Substitution	Relevant Trig Identity
$\sqrt{a^2-x^2}$	$x=a\sin(\theta), -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$	$1-\sin^2(\theta) = \cos^2(\theta)$
$\sqrt{a^2+x^2}$	$x=a\tan(\theta), -\frac{\pi}{2} < \theta < \frac{\pi}{2}$	$1+\tan^2(\theta) = \sec^2(\theta)$
$\sqrt{x^2-a^2}$	$x=a\sec(\theta), 0 \leq \theta < \frac{\pi}{2}$ or $\pi \leq \theta < \frac{3\pi}{2}$	$\sec^2(\theta) - 1 = \tan^2(\theta)$

Inverse substitution works as follows. If we write $x=g(t)$, then

$$\int f(x) dx = \int f(g(t)) g'(t) dt.$$

This is not the same as substitution. You can just apply inverse substitution to any integral directly--usually you get something even worse, but for the integrals in this section using a substitution can vastly improve the situation.

If $g$ is a $1-1$ function, then you can even use inverse substitution for a definite integral. The limits of integration are obtained as follows.

$$\int_{a}^{b} f(x) dx = \int_{g^{-1}(a)}^{g^{-1}(b)} f(g(t)) g'(t) dt.$$ (5.5)

To help you understand this, note that as $t$ varies from $g^{-1}(a)$ to $g^{-1}(b)$, the function $g(t)$ varies from $a=g(g^{-1}(a)$ to $b=g(g^{-1}(b))$, so $f$ is being integrated over exactly the same values. Note also that (5.3.2) once again illustrates Leibniz's brilliance in designing the notation for calculus.

Let's give it a shot with (5.3.1). From the table we use the inverse substition

$$x = \sec(\theta).$$

We get

$$\int_{\sqrt{2}}^2 \frac{1}{x^3 \sqrt{x^2-1}} dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{\sec(\theta)}{\sqrt{\sec^2(\theta)-1}} \sec(\theta)\tan(\theta) d\theta$$

$$= \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{\sec(\theta)}{\tan(\theta)} \sec(\theta)\tan(\theta) d\theta$$

$$= \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cos^(\theta) d\theta$$

$$= \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} 1 + \cos(2\theta) d\theta$$

$$= \frac{1}{2} \left[ \theta + \frac{1}{2}\sin(2\theta)\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}}$$

$$= \frac{\pi}{24} + \frac{\sqrt{3}}{8} - \frac{1}{4}$$

Wow! That was like magic. This is really an amazing technique. Let's use it again to find the area of an ellipse.

Example 5.3.1   Consider an ellipse with radii $a$ and $b$, so it goes through $(0,\pm b)$ and $(\pm a, 0)$. An equation for the part of an ellipse in the first quadrant is

$$y = b \sqrt{1-\frac{x^2}{a^2}} = \frac{b}{a}\sqrt{a^2-x^2}.$$

Thus the area of the entire ellipse is

$$A = 4 \int_{0}^a \frac{b}{a}\sqrt{a^2-x^2} dx.$$

The $4$ is because the integral computes $1/4$th of the area of the whole ellipse. So we need to compute

$$\int_{0}^a \sqrt{a^2-x^2} dx$$

Obvious substitution with $u=a^2-x^2$...? nope. Integration by parts...? nope.

Let's try inverse substitution. The table above suggests using $x = a\sin(\theta)$, so $dx = a\cos(\theta) d\theta$. We get

$$\int_{0}^{\frac{\pi}{2}} \sqrt{a^2 - a^2\sin^2(\theta)} d\theta = a^2 \int_{0}^{\frac{\pi}{2}} \cos^2(\theta) d\theta$$ (5.6)

$$= \frac{a^2}{2}\int_{0}^{\frac{\pi}{2}} 1+\cos(2\theta) d\theta$$ (5.7)

$$= \frac{a^2}{2}\left[ \theta + \frac{1}{2}\sin(2\theta) \right]_0^{\frac{\pi}{2}}$$ (5.8)

$$= \frac{a^2}{2}\cdot \frac{\pi}{2} = \frac{\pi a^2}{4}.$$ (5.9)

Thus the area is

$$4 \frac{b}{a} \frac{\pi a^2}{4} = \pi a b.$$

Consistency Check: If the ellipse is a circle, i.e., $a=b=r$, this is $\pi r^2$, which is a well-known formula for the area of a circle.

Remark 5.3.2   Trigonometric substitution is useful for functions that involve $\sqrt{a^2-x^2}$, $\sqrt{x^2+a^2}$, $\sqrt{x^2-a}$, but not all at once!. See the above table for how to do each.

One other important technique is to use completing the square.

Example 5.3.3   Compute $\int \sqrt{5 + 4x - x^2} dx$. We complete the square:

$$5 + 4x - x^2 = 5 - (x - 2)^2 + 4 = 9 - (x-2)^2.$$

Thus

$$\int \sqrt{5 + 4x - x^2} dx = \int \sqrt{9 - (x-2)^2} dx.$$

We do a usual substitution to get rid of the $x-2$. Let $u=x-2$, so $du=dx$. Then

$$\int \sqrt{9 - (x-2)^2} dx = \int \sqrt{9 - y^2} dy.$$

Now we have an integral that we can do; it's almost identical to the previous example, but with $a=9$ (and this is an indefinite integral). Let $y = 3\sin(\theta)$, so $dy = 3\cos(\theta)d\theta$. Then

$$\int \sqrt{9 - (x-2)^2} dx = \int \sqrt{9 - y^2} dy$$

$$= \int \sqrt{3^2 - 3^2\sin^2(\theta)} 3\cos(\theta) d\theta$$

$$= 9 \int \cos^2(\theta) d\theta$$

$$= \frac{9}{2} \int 1 + \cos(2\theta) d\theta$$

$$= \frac{9}{2} \left(\theta + \frac{1}{2}\sin(2\theta)\right) + c$$

Of course, we must transform back into a function in $x$, and that's a little tricky. Use that

$$x - 2 = y = 3\sin(\theta),$$

so that

$$\theta = \sin^{-1}\left(\frac{x-2}{3}\right).$$

$$\int \sqrt{9 - (x-2)^2} dx = \cdots$$

$$= \frac{9}{2} \left(\theta + \frac{1}{2}\sin(2\theta)\right) + c$$

$$=\frac{9}{2} \left[\sin^{-1}\left(\frac{x-2}{3}\right) + \sin(\theta)\cos(\theta) \right] + c$$

$$=\frac{9}{2} \left[\sin^{-1}\left(\frac{x-2}{3}\right) + \left(\frac{x-2}{3}\right) \cdot \left(\frac{\sqrt{9-(x-2)^2}}{3}\right) \right] + c.$$

Here we use that $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. Also, to compute $\cos(\sin^{-1}\left(\frac{x-2}{3}\right))$, we draw a right triangle with side lengths $x-2$ and $\sqrt{9-(x-2)^2}$, and hypotenuse $3$.

Example 5.3.4   Compute

$$\int \frac{1}{\sqrt{t^2-6t+13}} dt$$

To compute this, we complete the square, etc.

$$\int \frac{1}{\sqrt{t^2-6t+13}} dt = \int \frac{1}{\sqrt{(t-3)^2 + 4}} dt$$

[[Draw triangle with sides $2$ and $t-3$ and hypotenuse $\sqrt{(t-3)^2 + 4}$. Then

$$t-3 = 2\tan(\theta)$$

$$\sqrt{(t-3)^2 + 4} = 2\sec(\theta) = \frac{2}{\cos(\theta)}$$

$$dt = 2\sec^2(\theta) d\theta$$

Back to the integral, we have

$$\int \frac{1}{\sqrt{(t-3)^2 + 4}} dt = \int \frac{2\sec^2(\theta)}{2\sec(\theta)} d\theta$$

$$= \int \sec(\theta) d\theta$$

$$= \ln \vert \sec(\theta) + \tan(\theta)\vert + c$$

$$= \ln\left\vert \sqrt{(t-3)^2 + 4}{2} + \frac{t-3}{2}\right\vert + c.$$