# Thread: antiderivate of csc^3 x

1. Hey everybody,

I was just searching the Internet, trying to find an answer to the indefinite integral of cosecant cubed of x and I came across this forum (although I couldn't find an answer here). This looks like a really neat site.

So does anybody know the steps involved to getting the correct answer for this problem? (I can look up the answer in my book; I just don't know how to get there, and it has me stumped.)

Thanks.  2. ### Related Discussions:

• I'm going to let Demen Tolden, our resident integration expert-in-training, handle this one.  • Thanks! I am so happy that I get to use my new found integration skills to help someone else!

I'm working on it now though, just give me some time.   • Ok, this problem is done almost exactly like you would do
∫ sec<sup>3</sup> dx

Many calculus students seem to be using the same book as I am using, James Stewart's Calculus Early Transendentals. If this is the case there is an example that the book does in section 7.2 Trignometric Integrals, Example 8. The strategy, at least how I see it, is to set up your integral like this:
A = ∫ csc<sup>3</sup> dx

and then evaluate your integral so you also have to subract A on the right side, like so:

Integration by parts gives:
u = csc x
dv = csc<sup>2</sup>dx
du = -csc x cot x
v = -cot x

A = -cot x csc x - ∫ csc x cot<sup>2</sup>x

Now you may notice that if you use the trig identity
cot<sup>2</sup>x + 1 = csc<sup>2</sup>x

you can change this equation to this:
A = -cot x csc x - ∫ csc x ( csc<sup>2</sup> x - 1)dx

At this point I get excited and think "That's it! There is the other A I need contained in that second integrand!"
A = -cot x csc x - ∫ (csc<sup>3</sup> - csc x)dx
A = -cot x csc x - A + ∫ csc x dx
2A = -cot x csc x + ln (cscx - cot x)
A = (1/2)(-cot x csc x + ln (csc x - cot x) )

I am glad to be of assitance sir.   • Thanks so much! For some reason I thought I had to only use a u and du substitution, so didn't even try using integration by parts. That works much easier.

If you are willing (and want) to try and solve another one, I've got another for you: integral of du / (u sqrt (5-u^2)). It is in the chapter where I am supposed to be using a trigonometric substitution to get rid of the integral, but I can't seem to do that because of the 5. I tried doing various other substitutions, but can't find any that work.  • You are going to be running into problems like this very often. They will be in the form of
∫ dx / (9-x<sup>2</sup>)

or just
∫ (9 - x<sup>2</sup>)<sup>1/2</sup>

or
∫ dx / (x<sup>2</sup> + 6)

ect..

The pattern here is that the part of the integral that is difficult is a polynomial. The trick is to convert the polynomial into something that you can take the square root of such as sec<sup>2</sup> or sin<sup>2</sup>. If you have to deal with something such as
∫ dx / ( x ( 5 - x<sup>2</sup>)<sup>1/2</sup> )

You should find a substitution for x<sup>2</sup> where you can factor out the 5. In this case we could try to turn
5 - x<sup>2</sup>

into
5(1 - sin<sup>2</sup> Θ)

with just the right substitution for x. What I usually do is write:
x<sup>2</sup> = 5 sin<sup>2</sup> Θ

then solve for x:
x = 5<sup>1/2</sup> sin Θ

then we can find dx by differentiating since we have x alone
dx = 5<sup>1/2</sup> cos Θ dΘ

And now with this information, the integral becomes much easier to solve:
∫ 5<sup>1/2</sup> cos Θ (5<sup>1/2</sup> sin Θ 5<sup>1/2</sup> cos Θ)<sup>-1</sup> dΘ
∫ 5<sup>-1/2</sup> sec Θ dΘ
5<sup>-1/2</sup> ln (sec Θ + tan Θ)

and then you can just convert back to x since you know that sin Θ = x / (5<sup>1/2</sup>)  • Thanks!  • ∫ 5<sup>1/2</sup> cos Θ (5<sup>1/2</sup> sin Θ 5<sup>1/2</sup> cos Θ)<sup>-1</sup> dΘ
∫ 5<sup>-1/2</sup> sec Θ dΘ
5<sup>-1/2</sup> ln (sec Θ + tan Θ)
Oops. I think that should have been:
∫ 5<sup>-1/2</sup> csc Θ
5<sup>-1/2</sup> ln (csc Θ - cot Θ)

but the idea is still correct.  • Your answer is wrong. You don't have the constant of integration  • Call the math police!  • Demen

I think you are very brainy!!  • well you lose a lot of marks without it, sorry for pointing it out  • I wasn't objecting to you pointing out that the constant of integration was missing. Reminding people that antiderivatives are not unique is a worthwhile thing! Rather, I was reacting to you calling the answer "wrong" due this minor error. It's like you throwing your gin and tonic at me because I forgot the swizzle stick.  • Originally Posted by n77ler
well you lose a lot of marks without it, sorry for pointing it out
Why should you lose “a lot of” marks just for not remembering the arbitrary constant? One or two marks maybe, but surely not “a lot of”? If you’ve just done a difficult integration problem faultlessly except for the arbitrary constant at the end, surely you would expect to get at least 8 or 9 marks out of 10 for your hard work – but getting say 5 or 6 out of 10 just because you forgot a minor detail seems contrary to common sense!

The point about integration questions is to test how well students understand integration and can apply integration techniques (substitution, integration by parts, etc) – not how anal-retentive they are over minor details.  Bookmarks
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