Why Does Leg Selection in Trig Substitution Affect the Integral Result?

[FatalFlare]

I used trig substitution and got sqrt(x^2-9)+3*arcsin(3/x) which seems to be incorrect when I check it in my calculator and the textbook. I made a right triangle where one of the legs was sqrt(x^2-9) and it so happens that if you switch the leg the answer becomes sqrt(x^2-9) - 3*arctan(sqrt(x^2-9)/3)) which is the correct answer. It shouldn't matter which leg I choose but it does why?

 

[Mark44]

I moved this thread, which was originally posted in the HW section. It is not so much a homework problem as a question about why an answer can appear in two different forms. (Also, it might have drawn a warning in the HW section, as it was not posted using the homework template.)

 

[Ray Vickson]

I used trig substitution and got sqrt(x^2-9)+3*arcsin(3/x) which seems to be incorrect when I check it in my calculator and the textbook. I made a right triangle where one of the legs was sqrt(x^2-9) and it so happens that if you switch the leg the answer becomes sqrt(x^2-9) - 3*arctan(sqrt(x^2-9)/3)) which is the correct answer. It shouldn't matter which leg I choose but it does why?

Both $\sqrt{x^2-9}+3\arcsin(3/x)$ and $\sqrt{x^2-9} - 3\arctan(\sqrt{x^2-9}/3))$ have the same derivative, namely, your original integrand (assuming $x>3$).

 

[ssc]

Both $\sqrt{x^2-9}+3\arcsin(3/x)$ and $\sqrt{x^2-9} - 3\arctan(\sqrt{x^2-9}/3))$ have the same derivative, namely, your original integrand (assuming $x>3$).

But when we substitute something for x, why do they not give the same result? And also they don't have the same derivative..!? Only the derivate of sqrt(x^2-9)-3arctan(sqrt(x^2-9)/3) results in the original integrand

 

[PeroK]

But when we substitute something for x, why do they not give the same result? And also they don't have the same derivative..!? Only the derivate of sqrt(x^2-9)-3arctan(sqrt(x^2-9)/3) results in the original integrand

Please show your work!

 

[ssc]

Please show your work!

If we allow the x to b an integer like 4 that is greater than 3, sqrt(x^2-9)-3arctan(sqrt(x^2-9)/3) results in 0.477 while sqrt(x^2-9)+3arcsin(3/x) gives 5.18

 

[PeroK]

For $x > 3$, the functions differ by a constant, hence have the same derivative. Note that the first term in both functions is the same, so you can focus on the $\arcsin$ and $\arctan$ terms.

Note that for the derivative of the $\arctan$, in your final expression, $\frac{x^2 - 9}{9} + 1$ simpifies to $\frac{x^2}{9}$. And, you can see that the two derivatives are equal.

 

[PeroK]

PS the derivative of the $\arcsin$ function simplifies as well:
$$\frac{x}{\sqrt{x^2 - 9}} - \frac{9}{x^2\sqrt{1-\frac 9{x^2}}} = \frac{x}{\sqrt{x^2 - 9}} - \frac{9}{x\sqrt{x^2-9}} = \frac{x^2 - 9}{x\sqrt{x^2 - 9}} = \frac{\sqrt{x^2 - 9}}{x}$$Which is the same as the $\arctan$ derivative.

Remember that an antiderivative has an arbitary constant of integration, ususally denoted by $C$. In this case, the two functions differ by some constant (see your graph for $x > 3$): hence, have the same derivative. That wasn't so easy to see without doing some algebra on the resulting derivatives.

 

[PeroK]

@ssc Instead of using the triangle method, I would have tried the direct substitution:
$$x = 3\sec \theta = \frac 3 {\cos \theta}$$And used the standard identity:$$\sec^2\theta = \tan^2 \theta + 1$$Hence$$\sqrt{x^2 - 9} = \sqrt{9\sec^2\theta - 9} = \sqrt{9\tan^2\theta} = 3\tan \theta$$And $$dx = (\frac{3\sin\theta}{\cos^2 \theta})d\theta = (3\tan\theta \sec\theta)d\theta$$For the original integral, we have:
$$\int \frac{\sqrt{x^2 - 9}}{x} \ dx = \int \frac{3\tan \theta}{3\sec\theta}(3\tan\theta \sec\theta)d\theta$$$$ = 3\int \tan^2 \theta \ d\theta = 3\int (\sec^2 \theta - 1) \ d\theta$$$$= 3\tan\theta - 3\theta + C$$Note that the derivative of $\tan \theta$ is $\sec^2 \theta$, which is easy to show, but a very useful thing to remember!

Then, I would use the standard identity to finish this off:
$$= \sqrt{9 \tan^2 \theta} - 3\theta +C = \sqrt{9 \sec^2 \theta - 9} - 3\theta + C$$$$= \sqrt{x^2 - 9} - 3arcsec\big({\frac x 3}\big ) + C$$And we have yet another function that must have the same derivative!

In summary:
$$\int \frac{\sqrt{x^2 - 9}}{x} \ dx = \sqrt{x^2-9} - 3\arctan(\frac{\sqrt{x^2-9}}3)) + C_1 $$$$= \sqrt{x^2-9}+3\arcsin(\frac 3 x) + C_2$$$$= \sqrt{x^2 - 9} - 3arcsec\big({\frac x 3}\big ) + C_3$$
PS You can check those by differentiating them all. Or, you could find a list of inverse trig identities. Here are some, for example:
https://brilliant.org/wiki/inverse-trigonometric-identities/

You can see from that list that a lot of inverse trig functions are related by $\pm$ signs and constants. Which function you get for an integral may depend on the specific technique you use.

 

[PeroK]

PPS Let $x = \sec \theta$ for $x \ge 1$. So that $\theta = arcsec(x)$. Then:
$$\arctan(\sqrt{x^2 - 1}) = \arctan(\tan \theta) = \theta$$So, we have another inverse trig identity:
$$arcsec(x) = \arctan(\sqrt{x^2 - 1}) \ \ (x \ge 1)$$And we can use that to see that the first and third solutions above are indeed identical.

 

[PeroK]

PPS I think I understand the point of the triangle now. In this case, we have a triangle with adjacent side $3$ and opposite side $\sqrt{x^2 - 9}$ and a hypoteneuse of $x$.

We are free to pick any trig substitution:
$$\cos \theta = \frac 3 x, \ \sec \theta = \frac x 3, \ \sin \theta = \frac{\sqrt{x^2 - 9}}{x}, \tan \theta = \frac{\sqrt{x^2 - 9}}{3}$$They are all essentially the same in terms of the relationship between $x$ and $\theta$. It's really just a question of which one looks simplest. I picked $\sec \theta$ without drawing the triangle, because that looked the natural substitution to me.

Whatever you do, you should end up with the integral of $\tan^2 \theta$ and an intermediate answer of $3\tan \theta - 3\theta + C$.

At this point, you can then use any of the trig functions from the triangle to express $\theta$. I used $x = 3\sec \theta$, because that was the substitution I originally used. But, it's equally valid to note that we also know that $\tan \theta = \frac{\sqrt{x^2 - 9}}{3}$, and use the inverse of that.

This also explains why all the answers are valid. We can express $\theta$ in several different ways in terms of $x$. Just look at the triangle:
$$\theta = \arccos(\frac 3 x), \ \theta = arcsec(\frac x 3), \ \theta = \arcsin(\frac{\sqrt{x^2 - 9}}{x}), \ \theta = \arctan(\frac{\sqrt{x^2 - 9}}{3})$$This explains the point of drawing the triangle: you have all these different equations at your fingertips.

 

[MidgetDwarf]

PPS I think I understand the point of the triangle now. In this case, we have a triangle with adjacent side $3$ and opposite side $\sqrt{x^2 - 9}$ and a hypoteneuse of $x$.

We are free to pick any trig substitution:
$$\cos \theta = \frac 3 x, \ \sec \theta = \frac x 3, \ \sin \theta = \frac{\sqrt{x^2 - 9}}{x}, \tan \theta = \frac{\sqrt{x^2 - 9}}{3}$$They are all essentially the same in terms of the relationship between $x$ and $\theta$. It's really just a question of which one looks simplest. I picked $\sec \theta$ without drawing the triangle, because that looked the natural substitution to me.

Whatever you do, you should end up with the integral of $\tan^2 \theta$ and an intermediate answer of $3\tan \theta - 3\theta + C$.

At this point, you can then use any of the trig functions from the triangle to express $\theta$. I used $x = 3\sec \theta$, because that was the substitution I originally used. But, it's equally valid to note that we also know that $\tan \theta = \frac{\sqrt{x^2 - 9}}{3}$, and use the inverse of that.

This also explains why all the answers are valid. We can express $\theta$ in several different ways in terms of $x$. Just look at the triangle:
$$\cos \theta = \arccos(\frac 3 x), \ \theta = arcsec(\frac x 3), \ \theta = \arcsin(\frac{\sqrt{x^2 - 9}}{x}), \ \theta = \arctan(\frac{\sqrt{x^2 - 9}}{3})$$This explains the point of drawing the triangle: you have all these different equations at your fingertips.

Which I always found the triangle approach to be confusing to students. So i never taught it that way.

Instead I used the Unit Circle, and divided all terms by Sin^2 or Cos^2.

Down with the triangle!