Help finding the integral of 3x/sqrt(16 - x^4)

So I need to work out the integral of the expression '3x/sqrt(16 - x^4)' with respect to x.

My attempt at it was the following:

When integrating I took out the constant 3 in the numerator giving me 3 * integral x/sqrt(16 - x^4) dx

Next I used u-substitution where I let u = x^2 and dx = du/2x, substituting in those into the expression gets me '3 * integral 1/ 2* sqrt(16 - u^2) du

From that I can take out another constant which is just a half so now I have 3/2 * integral 1/sqrt(16 - u^2) du.

Im stuck up to this point now.

Can anyone please help what to do now or if I even made a mistake in the first steps so far.

Thanks!

Looks good. It's a standard integral which should be in your formula book to avoid workload.

Otherwise, notice that $16-u^2 = (4+u)(4-u)$ hence partial fractions are in order!

I'm not sure about the partial fractions here? Could you elaborate a bit more if you don't mind please?

There is a square root sign in my denominator, this is what I'm on so far ---> 3/2 * integral 1/ sqrt(16 - u^2) du

Ignore that bit. I didn't see the square root in your denominator.

Try a trig subsitution, if you don't want to use a formula booklet.

Okay so I heard the most suitable trig substitution for this type has to be letting u = 4 sin(theta) , then it implies that du= 4cos(theta) * d(theta)

So substituting in that I get 3/2 [4cos(theta)d(theta)/sqrt(16 - 16sin^2(theta)]

From the bottom of my denominator I have a constant 16 that I can factor out here so now I get the following:

3/2 [4cos(theta)d(theta)/sqrt (16 (1 - sin^2(theta)]

But the trig identity is cos^2x + sin^2x = 1

So therefore it implies that 3/2 [4cos(theta)d(theta) /sqrt(16 cos^2(theta)]

Again this can simplify down to ---> 3/2 [4cos(theta)d(theta) /4cos(theta)]

The 4cos(theta) cancels in the top and bottom of the fraction so that leaves me with: 3/2 * integral d(theta) = (theta) + c

But I know initially u = 4 sin(theta) so (theta) = sin^-1(u/4) as well as u being x^2

So to finish off I get the integral value to be 3/2 * arcsin(x^2/4) + c

Sorry if this took a little long to respond. I was writing and typing at the same time.

Is this the right answer?

https://www.wolframalpha.com/input/?i=int+3x%2Fsqrt(16+-+x%5E4)

Yep.

I've used the substitution x=2root(sinu) to begin with.

For the situation the the OP has, we can use another substitution : v = 4sinu

Otherwise, just the standard result as RDK suggested in post #3

Yes I have tried it and it seems to work now. Thanks!

FYI that sub comes from literally just combining the two subs you did.

First $u = x^2$

then $u = 4 \sin \theta$


Hence $x^2 = 4 \sin \theta \implies x = 2\sqrt{\sin \theta}$.

I'd be very surprised if he came up with it without thinking about these two subs individually first.

Ahh I see!

Just to clarify. Would take the general rule of thumb to always be that the trig sub is x = m*sin(theta), when the denominator we are looking is of the form of sqrt (m^2 - x^2)? Because the goal is to really break it down to a single trig function like you have mentioned before.

Would this always work when doing these types of integrals if the above conditions are met?

Yes (you mentioned the denominator of that form, but note that the numerator needs to be constant)

For $\displaystyle \int \dfrac{dx}{\sqrt{a^2 - x^2}}$ you sub in $x = a \sin \theta$.

Honestly, check your formula booklet. Do you have one? It should be in there without the need to remember!

It definitely is as I have looked into

https://qualifications.pearson.com/content/dam/pdf/A%20Level/Mathematics/2017/specification-and-sample-assesment/Pearson_Edexcel_A_Level_GCE_in_Mathematics_Formulae_Book.pdf

I do think it would be best to still commit it to memory, as it would just save the hassle of flicking through the pages during exams.

On that case it is safe to quote a result like that if it known to be a standard result without proof, right?

Okay so I heard the most suitable trig substitution for this type has to be letting u = 4 sin(theta) , then it implies that du= 4cos(theta) * d(theta)

So substituting in that I get 3/2 [4cos(theta)d(theta)/sqrt(16 - 16sin^2(theta)]

From the bottom of my denominator I have a constant 16 that I can factor out here so now I get the following:

3/2 [4cos(theta)d(theta)/sqrt (16 (1 - sin^2(theta)]

But the trig identity is cos^2x + sin^2x = 1

So therefore it implies that 3/2 [4cos(theta)d(theta) /sqrt(16 cos^2(theta)]

Again this can simplify down to ---> 3/2 [4cos(theta)d(theta) /4cos(theta)]

The 4cos(theta) cancels in the top and bottom of the fraction so that leaves me with: 3/2 * integral d(theta) = (theta) + c

But I know initially u = 4 sin(theta) so (theta) = sin^-1(u/4) as well as u being x^2

So to finish off I get the integral value to be 3/2 * arcsin(x^2/4) + c

Sorry if this took a little long to respond. I was writing and typing at the same time.

Is this the right answer?

yes it is