Recognizing When to Use Trig Substitution and When Not

[Jason76]

(a.)\(\displaystyle \int \dfrac{1 }{\sqrt{9 - x^{2}} }dx\) would require trig substitution. Yet,
(b.)\(\displaystyle \int \dfrac{x }{\sqrt{9 - x^{2}} }dx\) would not.

Let's look at (b.):

\(\displaystyle \int \dfrac{x }{\sqrt{9 - x^{2}} }dx\)

\(\displaystyle u = x^{2}\)


\(\displaystyle du = -2x\)

\(\displaystyle -\dfrac{1}{2} \int u^{-1/2} dx\)

\(\displaystyle -\dfrac{1}{2} \int \dfrac{u^{-1/2}}{-1/2} dx\)

What now?

 

[HallsofIvy]

Practice, Practice, Practice! The more problems you do the easier you will recognize what substitutions will work. If I had \(\displaystyle \int\frac{dx}{\sqrt{9- x^2}}\) I would think "\(\displaystyle cos^2(t)= 1- sin^2(t)\)" and use the substitution u= 3 sin(t). But looking at \(\displaystyle \int\frac{xdx}{\sqrt{9- x^2}}\), I would recognize the "x" in the numerator and think of "\(\displaystyle \frac{d(x^2)}{dx}= 2x\). But I wouldn't use "\(\displaystyle u= x^2\)"- I would substitute for the largest portion that I could. Here, I would use the substitution \(\displaystyle u= 9- x^2\) so that \(\displaystyle du= -2x dx\) and the integral becomes \(\displaystyle -\frac{1}{2}\int\frac{du}{\sqrt{u}}= \frac{1}{2}\int u^{1/2}du\). That appears to be what you did although you only write "\(\displaystyle u= x^2\). Allowing for that, you are fine until the last line. I have no idea where the "1/2" in the denominator of that last integral came from. To integrate \(\displaystyle \frac{1}{2}\int u^{1/2}du\) use the fact that the integral of \(\displaystyle u^n\) is \(\displaystyle \frac{1}{n+1}u^{n+1}+ C\).

 

[srmichael]

(a.)\(\displaystyle \int \dfrac{1 }{\sqrt{9 - x^{2}} }dx\) would require trig substitution. Yet,
(b.)\(\displaystyle \int \dfrac{x }{\sqrt{9 - x^{2}} }dx\) would not.

Let's look at (b.):

\(\displaystyle \int \dfrac{x }{\sqrt{9 - x^{2}} }dx\)

\(\displaystyle u = x^{2}\) ===> No. Let \(\displaystyle u=9-x^2\)


\(\displaystyle du = -2x\)

\(\displaystyle -\dfrac{1}{2} \int u^{-1/2} dx\)

\(\displaystyle -\dfrac{1}{2} \int \dfrac{u^{-1/2}}{-1/2} dx\) ===> No. \(\displaystyle -\dfrac{1}{2} \int u^{-1/2} dx=-\dfrac{1}{2}(\dfrac{u^{1/2}}{1/2})\) Now, substitute u back in and you're done (after you clean it up a little).



What now?

.

 

[Deleted member 4993]

(a.)\(\displaystyle \int \dfrac{1 }{\sqrt{9 - x^{2}} }dx\) would require trig substitution. Yet,
(b.)\(\displaystyle \int \dfrac{x }{\sqrt{9 - x^{2}} }dx\) would not.

Let's look at (b.):
?

We can use trig substitution for b also - I find it easier to solve the square-root problems with trig substitution.

\(\displaystyle \int \dfrac{x }{\sqrt{9 - x^{2}} }dx\) → x = 3 * sin(Θ) & dx = 3 * cos(Θ) dΘ

= \(\displaystyle \displaystyle\int \dfrac{3*sin(\theta)}{\sqrt{9 - 9*sin^{2}(\theta)} }3*cos(\theta) dΘ\)

= \(\displaystyle 3\displaystyle\int sin(\theta) dΘ\)

=-3 * cos(Θ) + C

Now substitute back.....

 

[Jason76]

We can use trig substitution for b also - I find it easier to solve the square-root problems with trig substitution.

\(\displaystyle \int \dfrac{x }{\sqrt{9 - x^{2}} }dx\) → x = 3 * sin(Θ) & dx = 3 * cos(Θ) dΘ

= \(\displaystyle \displaystyle\int \dfrac{3*sin(\theta)}{\sqrt{9 - 9*sin^{2}(\theta)} }3*cos(\theta) dΘ\)

= \(\displaystyle 3\displaystyle\int sin(\theta) dΘ\)

=-3 * cos(Θ) + C

Now substitute back.....

But You can't implement the quickie formula (bypassing trig substitution): \(\displaystyle \int \dfrac{du}{\sqrt{a^{2} - u^{2}}} = \sin^{-1} \dfrac{u}{a} + C\) in this case (problem (b)), cause there is no 1 on top.