# How does trigonometric substitution work?

I have read my book, watched the MIT lecture and read Paul's Online Notes (which was pretty much worthless, no explanations just examples) and I have no idea what is going on with this at all.

I understand that if I need to find something like $$\int \frac { \sqrt{9-x^2}}{x^2}dx$$

I can't use any other method except this one. What I do not get is pretty much everything else.

It is hard to visualize the bounds of the substitution that will keep it positive but I think that is something I can just memorize from a table.

So this is similar to u substitution except that I am not using a single variable but expressing x in the form of a trig function. How does this not change the value of the problem? To me it seems like it would, algebraically how is something like $$\int \frac { \sqrt{9-x^2}}{x^2}dx$$ the same as $$\int \frac {3\cos x}{9\sin^2 x}3\cos x \, dx$$

It feels like if I were to put in numbers for $x$ that it would be a different answer.

Anyways just assuming that works I really do not understand at all what happens next.

"Returning" to the original variable to me should just mean plugging back in what you had from before the substitution but for whatever unknown and unexplained reason this is not true. Even though on problems before I could just plug back in my substitution of $u = 2x$, $\sin2u = \sin4x$ that would work fine but for whatever reason no longer works.

I am not expected to do some pretty complex trigonometric manipulation with the use of a triangle which I do not follow at all, luckily though this process is not explained at all in my book so I think I am just suppose to memorize it.

Then when it gets time for the answer there is no explanation at all but out of nowhere inverse sin comes in for some reason.

$$\frac {- \sqrt{9-x^2}}{x} - \sin^{-1} (x/3) +c$$

I have no idea happened but neither does the author apparently since there is no explanation.

• Universal principle: don't use the same name for different things. Mar 4, 2014 at 7:47

It might help to not write $x$ for the before and after. That's misleading. What you're doing is writing $x=3\sin(\theta)$. You're inventing a new variable $\theta$ that relates to $x$ in a way that lets you use trig identities to solve your problem. Plugging in values after your transformation shouldn't give you the same values, since the new integral is in a different variable. We have: $$x=3\sin(\theta)$$ $x$ is a function of $\theta$. Differentiating we get: $$\frac{dx}{d\theta}=3\cos(\theta) \implies dx=3\cos(\theta) d\theta$$

So rewriting the integral in terms of our new variable, we get

$$\int \frac {\sqrt{9-9\sin^2(\theta)}}{9\sin^2(\theta)}\cdot3\cos(\theta)d\theta$$

Now we can use trig identities to evaluate this new integral and get rid of the troublesome square root, which is the reason we bothered with this substitution in the first place.

$$\int \frac {\sqrt{9\cos^2(\theta)}}{9\sin^2(\theta)}\cdot3\cos(\theta)d\theta$$ $$\int \frac {\cos^2(\theta)}{\sin^2(\theta)}d\theta$$ expanding the numerator as $1-\sin^2(\theta)$: $$\int \csc^2(\theta)-1 d\theta$$ evaluating, we get: $$-\cot(\theta)-\theta+C$$

But our integral was in terms of $x$. While this substitution was helpful, we're only halfway done, since we don't know what the expression looks like in our original variable. So now we need to sub in our original variable. $x=3\sin(\theta)$, so to get things in terms of $x$ we'd like to get as many $\sin(\theta)$s as we can. $\cot(\theta)=\frac{\cos(\theta)}{\sin(\theta)}=\frac{\sqrt{(1-\sin^2(\theta)}}{\sin(\theta)}$. By our substitution, $\sin(\theta)=x/3$. So we get $$-\frac{\sqrt{1-\frac {x^2} 9}}{\frac x 3}-\theta+C$$ We still have a $\theta$ we need to get rid of. To do this, we can just reverse our substitution. That is, $x=3\sin(\theta) \implies \theta=\sin^{-1}(x/3)$. C is still an arbitrary constant and so can stay. After simplifying the fraction, we get: $$-\frac{\sqrt{9-x^2}}{x}-\sin^{-1}(x/3)+C$$ Which is the answer. Hopefully this clears some things up.

• I completely lost track of what you are doing once you get to the $\int \frac {cos^2}{sin^2}$ part. To me it looks like that problem should just be $tan^2$ but you did something weird to it and made it into csc which I do not follow at all and I think you used a memorized formula to get cot?
– user138246
Jun 3, 2012 at 22:04
• Sorry. I did $\frac{cos^2}{sin^2}=\frac{1-sin^2}{sin^2}=\frac{1}{sin^2}-1=csc^2-1$. And yes I did use a memorized formula (i.e.experience with derivatives) to get $cot$. You can figure it out properly, but it's a big enough problem on its own that I figured doing it out would sidetrack too far from the actual question. Also $tan^2=\frac{sin^2}{cos^2}$, not $\frac{cos^2}{sin^2}$. Jun 3, 2012 at 22:11
• I still do not understand what is going on really, how is the subsitution like that legal? How is it not changing the value of everything? More importantly though how the heck does the arcsin stuff work? I don't see what happened. I don't even really know how arc sin works.
– user138246
Jun 3, 2012 at 22:26
• It is changing the value of everything. But afterwards I'm changing it back, so it doesn't matter. If I want to find the surface area of a cylinder, I solve the equivalent problem of the combined area of two circles and a rectangle. This is legal, if I can still make the cylinder out of those shapes afterwards. So I am changing how I represent the problem (in terms of a different variable), but because I can undo that transformation everything works out. $\arcsin$ is literally the inverse of $\sin$. No more, no less, just like how $\ln(x)$ is the reverse of $e^x$. Jun 3, 2012 at 22:55
• When we make this sort of substitution we are parametrising the variable X (excuse me if this isn't quite the right term). We change the same of the function, but as we change our limits (if we had any) and the dx to 3cos(θ)dθ we change the space we're in too. Relative to the space we work in, it's the same function, same question. We change it back as (whilst it's easier to solve) we want an answer in real space, not our imagined space. Jun 3, 2012 at 23:38

The following theorem is useful to keep in mind when making substitutions:

If function $x=\phi\left(t\right)$ satisfies the following conditions:

1) $\phi\left(t\right)$ is a continuous one-to-one function defined on the interval $\left[\alpha,\beta\right]$ and having a continuous derivative there.

2) values of $\phi\left(t\right)$ are contained within the interval $\left[a,b\right]$

3) $\phi\left(\alpha\right)=a$ and $\phi\left(\beta\right)=b$

Theorem. then for every function $f\left(x\right)$ continuous on $\left[a,b\right]$ the following formula for the change of variable in a definite integral holds:$$\int_{a}^{b}f\left(x\right)dx=\int_{\alpha}^{\beta}f\left[\phi\left(t\right)\right]\phi'\left(t\right)dt$$

• What does a circle with a line through it mean?
– user138246
Jun 3, 2012 at 22:05
• @Jordan Is the symbol for the greek letter "phi", that's all. Jun 3, 2012 at 22:10
• So it is a variable? I don't know why phi of t means. Is that a function? Like f(t)? So he is trying to say "If f(x) = f(t)" So that theorem is stating that if the two functions are equal then I have to change the bounds and then they are equal if I multply using the subsitution rule?
– user138246
Jun 3, 2012 at 22:11
• It's just the name of a function, @Jordan...just like writing $\,f(x)\,,\,g(x)\,$ or so...but it's another function. Jun 3, 2012 at 22:19
• @DonAntonio I guess I will translate what he wrote into terms I can understand. So the first thing I need to do is verify that the function is one to one. I do not know of any easy way to do this, maybe just make sure it isn't an absolute value or something like that? Then I need to make sure that the values of the function are in te interval. I don't really know what that means either, so I have to test each point or the end points? Then 3) is really weird and I have no idea at all what that means.
– user138246
Jun 3, 2012 at 22:29

You must use a name other than $x$ when you substitute, maybe say let $x=3\sin t$. (It makes no sense to let $x=3\sin x$.) Using $x$ for the substitution is probably one of the big reasons for being confused about the process.

We have then $dx=3\cos t\,dt$, and $\sqrt{9-x^2}=\sqrt{9(1-\sin^2 t}=3\cos t$. So we arrive at $$\int \frac{\cos^2 t}{\sin^2 t}\,dt.$$ To evaluate, use $\cos^2 t=1-\sin^2 t$. So we want $\int \frac{1-\sin^2 t }{\sin^2 t}\,dt$.

This simplifies to $\int \left(\frac{1}{\sin^2 t} -1\right)\,dt$, and then to $\int(\csc^2 t-1)\,dt$.

Now it is useful to remember that $-\cot t$ is an antiderivative of $\csc^2 t$. We conclude that our integral is equal to $$-\cot t + t+C.$$ Finally, we want to go back to $x$. Remember that $x=3\sin t$, so $\sin t=x/3$, and therefore $t=\sin^{-1}(x/3)$.

To express $\cot t$ in terms of $x$, use the fact that $\frac{\cos t}{\sin t}$. We have $\sin t=x/3$, and $\cos t=\sqrt{1-(x/3)^2}=\frac{1}{3}\sqrt{9-x^2}$. When we put the pieces together we get the expression that you quoted.

• I do not understand how you can just subsitute something in for x without changing the problem. I see how using 3cost is wanted because then when you square it you can factor out a 9 of the radicand and get the form of a trig identity. I do not follow at all what happens in the first line of your calculation or why you have $cos^2 t/sin^ 2 t$
– user138246
Jun 3, 2012 at 22:00
• Just like we temporarily let $u=x^2$ when differentiating, for instance, $\cos(x^2)$. When we substitute we get $3\cos t\,dt$ from $dx$, and $3\cos t$ from the square root term. Then we get $9\sin^2 t$ at the bottom from the $x^2$ at the bottom. Jun 3, 2012 at 22:07
• Maybe I am thinking of this in the wrong way but this is basically an equation isn't is? I am trying to find the integral of something so why am I only manipulating one side? To me it seems like I am just changing the problem./
– user138246
Jun 3, 2012 at 22:11
• Well, if you have a good guess as to the integral, you can check it by differentiating. That's usually a good idea anyway. Jun 3, 2012 at 22:13
• @Jordan while you might be worried that you are changing the whole integral when you substitute $x$ with a trig function it is only temporarily and you will undo the substitution later. When you write $x=\sin u$ you will substitute $u=\arcsin x$ later. So essentially what you are writing is $x=\sin (\arcsin (x) )= x$. Note that the $\sin$ and $\arcsin$ undo each other. Therefore you are just exploiting the properties of trig functions to simplify the integrand before you substitute back at the end.
– E.O.
Jun 3, 2012 at 23:17

Imagine this. Drop your example for a sec. Is there any way that you can write an equation where $\sin{a} = \sqrt{1-x^{2}}$ ? Plug in an x, and no matter what the x is, you can find an a that satisfies the equation? If so, great! There is only one rule: the two functions must have the same range, or must output the same area of y values, and we use a different variable. For example, sin(a) only outputs numbers between -1 and 1, and the same goes for our second equation. Also, it doesn't use the same variable, so we're good!

If you don't believe me, let's solve for a. Taking the arcsin yields $a = \arcsin{ \sqrt{1-x^{2}}}$. If you pick any value of x, you will get a value fora. If you solved for x, you would get $\sqrt{1-\sin^{2}{a}}=x$.

The important thing to recognize is that we are just finding new functions to represent the same number.

This is all just a different way of looking at the same number.

The new variable ensures a new function, and the equivalent range ensures the new function can still produce the same numbers.

In the equation you give, any x value greater than or equal to 3 would put a negative number under the radical. This is also true for any number less than -3, because squaring gets rid of the negative. This means the domain is all real numbers greater than -3 but less than 3. How convenient! That entire domain is also found in 3 cos(theta), which oscillates between -3 and 3. So they have the same range! Perfect!

Try imagining this a different way. Let's say you want to solve the equation I gave, $\int x^{4} dx$. This is very easy, but sometimes you have to make the easy things hard to make the hard things easy. Let's say that you want to substitute in $x=4y$ and the necessary $dx = dy$. Can you do it? Well, why wouldn't you be able to? They output the same range: real numbers. They have different variables. So let's do it!

We now have

$\int \left (4y \right )^{4} dy$

which is just

$\int \ 256y^{5} dy$ . Completely pointless but bear with me. This is just equal to

$\frac{256y^{5}}{5}$

Now let's plug back in x. Our original equation mentioned that $x=y^{2}$. Solving for y we have $y=\frac{x}{4}$. By putting that in our new equation we have our new equation as

$\frac{x^{5}}{5}$. Seem familiar? We have seen that the only two things necessary to do a substitution is that

a. you have the same range in both and

b. you use a different variable.

3 cos(theta) = x is just the same. In the equation referenced you have a range between -3 and 3, as does 3 cos(theta). Obviously, they use different variables, so you are good.

• Good answer, +1. Aug 27, 2020 at 21:05

What you did is a substitution: $x=3\sin u$ so $u=\arcsin\left(\frac{x}{3}\right)$, this is what you use to substitute back for the final answer. Just to complete the substitution, you use $dx=3\cos u \,du$: $$\int \frac { \sqrt{9-x^2}}{x^2}dx=\int\frac{\sqrt{9-9\sin^2u}}{9\sin^2u}3\cos u \,du=\int\frac{3\cos u}{9\sin^2u}3\cos u \,du=\int\frac{1-\sin^2 u}{\sin^2u} \,du=\int\left(\frac{1}{\sin^2u}-1\right) \,du=-\cot u-u+C$$ Now you have to return to a function of $x$. For that you substitute: $u=\arcsin\left(\frac{x}{3}\right)$ $$-\cot u-u+C=-\cot\arcsin\left(\frac{x}{3}\right)-\arcsin\left(\frac{x}{3}\right)+C$$

• So $x=3\sin u$ so $u=\arcsin\left(\frac{x}{3}\right)$ are equal statements? I don't quite understand arcsin and such.
– user138246
Jun 3, 2012 at 21:56

There are some basic trigonometric identities which is not hard to memorise, one of the easiest and most important being $\,\,\cos^2x+\sin^2x=1\,\,$ , also known as the Trigonometric Pythagoras Theorem.

From here we get $\,1-\sin^2x=\cos^2x\,$ , so (watch the algebra!)$$\sqrt{9-x^2}=\sqrt{9\left(1-\left(\frac{x}{3}\right)^2\right)}=3\sqrt{1-\left(\frac{x}{3}\right)^2}$$From here, the substitution proposed for the integral is $$\displaystyle{\sin\theta=\frac{x}{3}\Longrightarrow \cos\theta\, d\theta=\frac{1}{3}dx}\,\,,\,x=3\sin\theta\,,\,dx=3\cos\theta\,d\theta$$ so in the integral we get$$\int \frac{\sqrt{9-x^2}}{x^2}\,dx =\int \frac{3\sqrt{1-\left(\frac{x}{3}\right)^2}}{x^2}\to\int\frac{\rlap{/}{3}\sqrt{1-\sin^2\theta}}{\rlap{/}{9}\sin^2\theta}\,\rlap{/}{3}\cos\theta\,d\theta=$$$$\int\frac{\cos\theta\,\cos\theta}{\sin^2\theta}\,d\theta$$which is what you have in your book...:)

• I lost track of the math in the first step, I do not see how they are equivelant statements.
– user138246
Jun 3, 2012 at 22:15
• @Jordan: Just multiply $9 - x^2$ by $\frac{9}{9}$ (that is, multiply it by 9 and divide it by 9), which is equivalent to multiplying by 1, like this: $9-x^2 = 9\times\frac{\left(9-x^2\right)}{9} = 9\left(1-\frac{x^2}{9}\right) = 9\left(1-\left(\frac{x}{3}\right)^2\right)$.
– favq
Jun 10, 2012 at 20:22