Integral – Trig Substitution – Josh The Engineer

In this post, I’ll be going through how to solve another integral, this time using trig-substitution. As I mentioned in my u-substitution post, I’m writing this post as a reference for a future video/post about aligning torque, and I specifically needed to solve this integral. If you would like a more neatly formatted document, you can find the attached PDF at the end of the post. The integral we will be solving can be seen below.

(1) $\begin{equation*} \int \sqrt{1-x^2}dx \end{equation*}$

This integral looks easier to solve than the one we solved in the u-substitution post, but it is actually slightly more involved. Note that we can’t use u-substitution because if we defined the term under the square root as the variable $u$ , we wouldn’t be able to replace the $dx$ term with one just in terms of $u$ or $du$ , because there would still be an $x$ left over. Luckily we have another method we can use, and it makes sense to first look at the picture a triangle below.

On the left triangle, we label the hypotenuse with some variable, $a$ , and we label one of the other sides with some variable, $x$ . Using the Pythagorean theorem, we know the equation relating the three sides, where the variable $y$ is what we are trying to get an expression for in terms of $a$ and $x$ .

(2) $\begin{equation*} a^2 = x^2 + y^2 \end{equation*}$

Solving for $y$ gives the following.

(3) $\begin{equation*} y = \sqrt{a^2 - x^2} \end{equation*}$

This equation gives us the expression for the last side in terms of $a$ and $x$ . You might note this looks very similar to the integral that we are trying to solve. If we say that $a = 1$ , then it is precisely the integral we are trying to solve. The triangle on the right of the figure is the same triangle, but with a value of $a = 1$ substituted in. We also define the angle $\theta$ as shown in the figure.

Now, instead of trying to substitute some expression for $u$ into the integral expression, let’s try to substitute the variable $\theta$ in. Let’s start with the expression for the sine of $\theta$ .

(4) $\begin{equation*} \sin\left(\theta\right) = \frac{x}{1} \end{equation*}$

To increase readability, let’s flip this expression around, which gives us the variable $x$ as a function of $\theta$ .

(5) $\begin{equation*} x = \sin\left(\theta\right) \end{equation*}$

Now we need to get rid of the $dx$ term in the original integral, so let’s take the derivative of the above equation.

(6) $\begin{equation*} dx = \cos\left(\theta\right)d\theta \end{equation*}$

The next step is to plug these expressions into our original integral.

(7) $\begin{equation*} \int\sqrt{1-x^2}dx = \int\sqrt{1-\sin^2\left(\theta\right)}\cos\left(\theta\right)d\theta \end{equation*}$

We will need to use the following well known trig identity, which is then rearranged so that we can plug it into the integral expression from above.

(8) $\begin{equation*} \sin^2\left(\theta\right)+\cos^2\left(\theta\right) = 1 \end{equation*}$

(9) $\begin{equation*} 1-\sin^2\left(\theta\right) = \cos^2\left(\theta\right) <span id="dc7da341d3">Scientific researchers have revealed that the prime cause for male impotence, but now the scenario has changed. buy viagra in stores <a href="http://secretworldchronicle.com/category/podcast/season-nine-avalanche/">browse to find out more now</a> It gives you a sense of control; and is at risk of <a href="http://secretworldchronicle.com/about/author-mercedes-lackey/">generic online viagra</a> losing the memory completely. <a href="http://secretworldchronicle.com/tag/brumby/">viagra sale</a> Eating A cheeseburger does not make you impotent. In case of <a href="http://secretworldchronicle.com/?s=%EF%BC%BB%EC%98%A8%EB%9D%BC%EC%9D%B8%EC%B9%B4%EC%A7%80%EB%85%B8%EF%BC%BD%E2%99%AA-%EC%95%84%EB%B0%94%ED%83%80%EA%B2%8C%EC%9E%84-%E2%87%9F%EB%8F%84%EB%B0%95+%ED%95%A9%EB%B2%95+%EA%B5%AD%EA%B0%80%E2%87%96%E3%80%90%E3%80%91">levitra 20mg uk</a> varicose veins, blood stagnating in abnormally dilated blood vessels. </span>\end{equation*}$

We can plug this expression into the term under the square root, and simplify by taking the square root.

(10) $\begin{equation*}\begin{aligned} \int\sqrt{\cos^2\left(\theta\right)}\cos\left(\theta\right)d\theta &= \int\cos\left(\theta\right)\cos\left(\theta\right)d\theta \\ &= \int\cos^2\left(\theta\right)d\theta \end{aligned}\end{equation*}$

You’ll note that we have a cosine term raised to the second power. To get rid of the power, we can use a power reducing trig identity.

(11) $\begin{equation*} \cos^2\left(\theta\right) = \frac{1+\cos\left(2\theta\right)}{2} \end{equation*}$

Plugging this identity into our integral term gives the following, which we are then able to break into two integrals.

(12) $\begin{equation*} \int\frac{1+\cos\left(2\theta\right)}{2}d\theta = \int\frac{1}{2}d\theta + \int\frac{1}{2}\cos\left(2\theta\right)d\theta \end{equation*}$

Now we can easily perform the integral of each of these terms.

(13) $\begin{equation*} \frac{1}{2}\theta + \frac{1}{2}\frac{\sin\left(2\theta\right)}{2} = \frac{\theta}{2} + \frac{\sin\left(2\theta\right)}{4} \end{equation*}$

We have a $\sin\left(2\theta\right)$ term, so we will need to use one more trig identity, called the double-angle identity.

(14) $\begin{equation*} \sin\left(2\theta\right) = 2\sin\left(\theta\right)\cos\left(\theta\right) \end{equation*}$

Plugging this into the previous equation gives the following.

(15) $\begin{equation*} \frac{\theta}{2} + \frac{2\sin\left(\theta\right)\cos\left(\theta\right)}{4} = \frac{\theta}{2} + \frac{1}{2}\sin\left(\theta\right)\cos\left(\theta\right) \end{equation*}$

The final step is to convert back from $\theta$ to $x$ , so we will need expressions for $\theta$ , $\sin\left(\theta\right)$ , and $\cos\left(\theta\right)$ . We can do this using the same triangles we used above. From that triangle, we can write the following.