Trigonometric substitution (sine)

Substitution using sine function to simplify integrals

This is a technique where we substitute our integration variable by a trigonometric function. It is particularly useful for simplifying integrals involving radicals. In this lecture we focus on only the sine function.

Trigonometric substitution: the basic idea

This is a useful technique where we substitute our integration variable by a trigonometric function (which is the opposite from what we learned earlier). This technique is particularly useful for simplifying integrals involving radicals (square root, etc.).

The foundation of this idea is the Pythagorean theorem: \[ (\sin \theta)^2 + (\cos \theta)^2 = 1, \] which means \[ (\cos \theta)^2 = 1- (\sin \theta)^2. \] This is how we get rid of square roots.

There are, of course, many variations on this theme. We will start with a simple example to illustrate the basic idea. For simplicity, we will only discuss the "substitution" part of this idea, and the question of how to compute the resulting integrals are covered in other lectures.

Compute \[ \int \sqrt{1 - x^2} \, dx.\]

Notice that since $\sqrt{1-x^2}$ is the integrand, it is necessary that $1 - x^2 \ge 0$, i.e., $-1 \le x \le 1$.

This happens to be the range of the sine function. Therefore, we do not loose anything if we introduce a new variable $\theta$...

and let $\displaystyle \color{red}{x = \sin(\theta)}$. Then $\displaystyle \theta = \sin^{-1}(x)$ and $\displaystyle \color{blue}{dx = \cos(\theta) d\theta}$.

It is important to note that, since $\theta = \sin^{-1}(x)$, we are implicitly requiring $-\pi/2 \le \theta \le \pi/2$. With this substitution, we have \[ \sqrt{1 - x^2} = \sqrt{1 - (\sin \theta)^2} = \sqrt{ (\cos \theta)^2 } = | \cos \theta | = \cos \theta. \]

Here, can safely drop the absolute value because $\pi/2 \le \theta \le \pi/2$.

Then $\displaystyle \int \color{purple}{\sqrt{1 - x^2}} \,\, \color{blue}{dx}$ $\displaystyle = \int \color{purple}{\cos \theta} \, \color{blue}{ \cos(\theta) \, d\theta}$ $\displaystyle = \int [\cos(\theta)]^2 d\theta$

which is an integral that we can actually compute. (We have discussed how to compute such integrals in other lectures)

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

Just like in the previous example, $\sqrt{1-x^2}$ is in the integrand, so $1 - x^2 \ge 0$, i.e., $-1 \le x \le 1$.

Therefore, we do not loose anything if we introduce a new variable $\theta$ and...

let $\displaystyle \color{red}{x = \sin(\theta)}$. Then $\displaystyle \theta = \sin^{-1}(x)$ and $\displaystyle \color{blue}{dx = \cos(\theta) d\theta}$.

Notice that since $\theta = \sin^{-1}(x)$, we are requiring $-\pi/2 \le \theta \le \pi/2$.

Then $\displaystyle \sqrt{1 - \color{red}{x}^2} = \sqrt{1 - (\color{red}{\sin \theta})^2}$ $\displaystyle = \sqrt{ (\cos \theta)^2 }$ $\displaystyle = | \cos \theta |$ $\displaystyle = \cos \theta$

Therefore, the original integral is turned into \[ \int \frac{ {\color{red}{x} }^2}{ \color{purple}{ \sqrt{1 - x^2} } } { \color{blue}{dx} } = \int \frac{ (\color{red}{\sin \theta})^2}{ \color{purple}{\cos \theta} } \color{blue}{ \cos \theta \, d\theta} = \int (\sin \theta)^2 d\theta, \]

which is an integral that we can compute. Again, for simplicity, we only focus on the substitution part, and the actual computation of this resulting integral is covered in other lectures.

Summary of the main idea

In the two examples on the previous pages, we substitute our integration variable by a sine function. This is particularly useful for simplifying integrals involving radicals of the form $\sqrt{1 - x^2}$.

We used the substitution $\color{red}{x = \sin(\theta)}$ and $\color{blue}{dx = \cos(\theta) \, d\theta}$.

Then following the Pythagorean theorem $(\sin \theta)^2 + (\cos \theta)^2 = 1$, i.e., $(\cos \theta)^2 = 1- (\sin \theta)^2$. We can get rid of the square root.

In particular, \[ \sqrt{1 - x^2} = \sqrt{1 - (\sin \theta)^2} = \sqrt{ (\cos \theta)^2 } = | \cos \theta | = \cos \theta. \]

Therefore, \[ \int f \left( \color{purple}{ \sqrt{1-x^2} }, \color{red}{x} \right) \, \color{blue}{dx} = \int f \left( \color{purple}{ \cos(\theta) }, \color{red}{\sin(\theta)} \right) \, \color{blue}{\cos(\theta) \, d\theta} \] for any function $f$. This is the main idea.

Slightly more general situations

This technique can simplify integrals involving $\sqrt{a - b x^2}$ for $a,b > 0$.

Compute \[ \int \sqrt{4 - 3x^2} \, dx.\]

Instead of $1 - x^2$, we have $4-3x^2$ inside the square root, but we can easily compensate for this.

$\displaystyle \int \sqrt{4 - 3x^2} \, dx$ $\displaystyle = \int \sqrt{\color{red}{4} \cdot 1 - \color{red}{4} \cdot \frac{3}{4} x^2} \, dx$ $\displaystyle = \int \sqrt{\color{red}{4} \left( 1 - \frac{3}{4} x^2 \right)} \, dx.$

Therefore, we need $x^2$ to be $\frac{4}{3} \sin^2(\theta) = \left( \frac{2}{ \sqrt{3} } \sin \theta \right)^2$.

I.e., $\displaystyle \color{red}{x = \frac{2}{ \sqrt{3} } \sin \theta}$, then $\displaystyle \theta = \sin^{-1} \frac{\sqrt{3}x}{2}$, and $\displaystyle \color{blue}{dx = \frac{2}{\sqrt{3}} \cos \theta \, d\theta}$.

Then the original integral becomes

$\displaystyle \int \sqrt{4 - 3 \left(\color{red}{\frac{2}{ \sqrt{3} } \sin\theta} \right)^2} \color{blue}{\frac{2}{\sqrt{3}} \cos \theta \, d\theta}$ $\displaystyle = \int 2\sqrt{1 - (\sin\theta)^2} \frac{2}{\sqrt{3}} \cos \theta \, d\theta$

$\displaystyle = \int \frac{4}{\sqrt{3}} \sqrt{(\cos \theta)^2} \, \cos\theta \, d\theta$ $\displaystyle = \int \frac{4}{\sqrt{3}} \cos \theta \, \cos\theta \, d\theta$ $\displaystyle = \frac{4}{\sqrt{3}} \int (\cos \theta)^2 d\theta$

Summary of the general form

In the previous example, we saw how the idea of substitution using the sine function work in general. This can be applied to simplify integrals involving radicals of the form $\sqrt{a - b x^2}$ for any $a,b > 0$.

We used the substitution $\color{red}{x = \frac{\sqrt{a}}{\sqrt{b}} \sin(\theta)}$ and $\color{blue}{dx = \frac{\sqrt{a}}{\sqrt{b}} \cos(\theta) \, d\theta}$.

Then following the Pythagorean theorem, we can get rid of the square root.

$\displaystyle \sqrt{a - b \color{red}{x}^2} = \sqrt{a - b \left( \color{red}{ \frac{\sqrt{a}}{\sqrt{b}} \sin \theta} \right)^2} $ $\displaystyle = \sqrt{a - a (\sin \theta)^2} $ $\displaystyle = \sqrt{a}\sqrt{1 - (\sin \theta)^2} $

$\displaystyle = \sqrt{a} \sqrt{ (\cos \theta)^2 } $ $\displaystyle = \sqrt{a} | \cos \theta | $ $\displaystyle = \sqrt{a} \cos \theta. $

Therefore, \[ \int f \left( \color{purple}{ \sqrt{a- bx^2} }, \color{red}{x} \right) \, \color{blue}{dx} = \int f \left( \color{purple}{ \sqrt{a} \cos(\theta) }, \color{red}{\frac{\sqrt{a}}{\sqrt{b}} \sin(\theta)} \right) \, \color{blue}{\frac{\sqrt{a}}{\sqrt{b}} \cos(\theta) \, d\theta}. \]