# Integrating trig substitution triangle equivalence

When we integrate certain integrals, such as

$$\int \frac{x^2}{\sqrt{16-x^2}} dx$$

We can make a substitution like $x = 4 \sin \theta$

Then we can simplify the above integral to the following: $$8 \theta - 8 \sin \theta \cos \theta + C$$

I then learned we can use a right angled triangle to find alternate expressions for $\frac{x}{4} = \sin \theta$ such as $\frac{\sqrt{16-x^2}}{4} = \cos \theta$ and substitute theta to find the answer $8 \arcsin \frac{x}{4} - \frac{x}{2} \sqrt{16-x^2} + C$

But clearly when I graph the two functions

$$y=\arcsin \left(\frac{x}{4}\right)$$

and

$$y=\arccos \left(\frac{\sqrt{16-x^2}}{4}\right)$$

They are only equal for $x \ge 0$ according to https://www.desmos.com/calculator

Whats going on here? Why does this work? Why can we make this equivalent triangle substitution when the functions clearly arent equal to each other on $x < 0$?

• I think when you substitute $\theta$ for $x$ in the integral, the square root in the denominator would evaluate to two different values depending upon the domain of values of $\theta$. These two values are $4cos\theta$ or $-4cos\theta$. So, the original integral would be split into sum of two integrals - the denominator in each being one of these two values. Also, the validity of such substitution is circumspect, since $4 sin \theta$ is bounded between $-4$ and $4$ whereas the original integral is indefinite – Deepak Gupta Sep 2 '15 at 9:04
• Edit: the substitution is valid since $16-x^2>0$ , so please ignore the latter part of my previous comment (this is to recant my earlier skepticism upon the validity of your substitution) – Deepak Gupta Sep 2 '15 at 9:24

Read your solution carefully. You make a triangle and use it to get the expression. Now you only tell me, how can the length of a side of a triangle be negative?
Now if you want them to be equal for every value. Then observe that $\cos\theta=\pm\frac{\sqrt{16-x^2}}{4}$ due to the square rooting done.

• So when we use this triangle method, the domain has another restriction added, $x \gt 0$ ? Even if the domain beforehand was say $x \in [-a, a]$ it is now $x \in [0, a]$ ? – Jason Sep 2 '15 at 11:22
• BTW, see, they both indeed are equal for $-2$. – Aditya Agarwal Sep 2 '15 at 11:23
• See your real fault in the edit. You have ignored the minus part in square rooting, I don't think you require the $Arccos$ relation, because when we solve integrals, we deal only in functions. – Aditya Agarwal Sep 2 '15 at 11:25
• but isnt $\cos \theta \ge 0$ in the defined domain of $\theta$ for our sine substitution? – Jason Sep 2 '15 at 11:47
• Why?. Yes in the triangle it is. That is what my solution says. But the latter part in my solution says that if you want to know that how to get your expression true for all $x$, then use this method. – Aditya Agarwal Sep 2 '15 at 13:22

Let $y=\arcsin\dfrac x4\implies-\dfrac\pi2\le y\le\dfrac\pi2\implies\cos y\ge0$

and $\sin y=\dfrac x4,\cos y=+\dfrac{\sqrt{16-x^2}}4$

and $dy=\dfrac{dx}{\sqrt{16-x^2}}$

$$\int\dfrac{x^2}{\sqrt{16-x^2}}dx=\int(4\sin y)^2\ dy=8\int(1-\cos2y)dy$$

$$=8y-4\sin2y+k$$

$$=8\arcsin\dfrac x4-8\cdot\dfrac x4\cdot\dfrac{\sqrt{16-x^2}}4+k$$

• I'm confused because the domain for y is $y\ in[-\frac{\pi}{2}, \frac{\pi}{2}]$ but the two functions $y= \arcsin \frac{x}{4}$ and $y = \arccos \frac{\sqrt{16-x^2}}{4}$ are only equal on $y \in [0, \frac{pi}{2}]$ – Jason Sep 2 '15 at 11:39
• EDIT: Sorry, in my above comment when I talk about domain for y, I really mean x. The two functions are only equal on the domain $x \ge 0$ So does this mean our answers domain is $x \ge 0$ ? – Jason Sep 2 '15 at 12:12
Use Arccos relation instead of the arccos function. Note that Arccos is a relation and not a function defined as "set of all values whose cosine is the domain of the Arccos relation" $$y=Arccos \left(\frac{\sqrt{16-x^2}}{4}\right)$$ This relation generates infinite possibilities for 'y'. This set of solutions, let's call it set A
Likewise, use Arcsin relation instead of of the arcsin function. $$y=Arcsin \left(\frac{x}{4}\right)$$ This relation also generates infinite possibilities for 'y' This set of solutions, let's call it set B
Your solution set for 'y' is $A \bigcap B$