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Use an inverse trig substitution on the integral...

Mathematics
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|dw:1358909488344:dw|
For inverse trig function, you want to get it into a form that looks like arctan, arcsin, arcos, etc., The first step is to complete the square.
complete the square in the denominator... okkkk

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you're going to have to refresh my memory on how to do that :/
|dw:1358910770499:dw|
\[ax^2+bx+c \] \[(\frac{ b }{ 2 })^2 \]
i'm sorry, i forget what to do...
divide by two and square the answer of the 2nd term (b).
then add it to C.
|dw:1358910964480:dw|
Yeah. Now, you can factor it to get something that looks like an inverse trig, so that you get: \[(x \pm a)^2 + 1\] where you will tell what inverse trig function you have. Try it.
Essentially, you might need to do more work, like factor out a number or two. But remember that you have it in a square root! so I think it might be arcos? Just a guess. It wont be arctan since arctan does;t have a square root.
|dw:1358914184400:dw|@abb0t
Now use u-substitution.
of what?
u - x-1 ?
yeah
u = x-1 ***
i don't see where that takes us.
|dw:1358914405152:dw|
Let's not do a U-sub, that seems rather unnecessary. We'll already be doing a trig sub, so it's better if we can limit the number of substitutions that we apply. Your factoring looks good. So we're at this point.\[\large \int\limits\frac{dx}{\left[2(x-1)^2-3\right]^{3/2}}\]
Ok this next part is going to be a little tricky to understand, let me see if I can put it into the right words. There are 3 primary forms that we look for when making a trig sub. We'll look at the one that specifically applies to this problem. Ignore the 2 coefficient on X in the bottom for a moment. Our intention is to get rid of the addition or subtraction in the bottom. If we can do that, simplify it down to ONE term, we can easily apply the exponent on the outside to it. If we see:\[\large x^2-a^2\]We want to let \(\large x=a\sec \theta\) That will do this for us,\[\large x^2-a^2 \qquad \rightarrow \qquad a^2\sec^2 \theta-a^2 \qquad \rightarrow \qquad a^2(\sec^2\theta-1)\] From there, we can recall that \(\large \sec^2\theta-1=tan^2\theta\) and make the switch. \[\large a^2(\tan^2\theta)\] Now you might be saying, "But zep! We don't have an a^2! We just have a 3!" Well baldy this is where it gets a little tricky. We need to get that ugly 2 off of our x term.\[\large \int\limits\limits\frac{dx}{\left[2\left((x-1)^2-\dfrac{3}{2}\right)\right]^{3/2}}\]
"But zep! We still don't have a nice \(a\) value!" Well that's just the way it works out sometimes. We DO have an \(a\), it's just a little ugly :D So in our case, our \(a\) value is going to be \(\dfrac{\sqrt3}{\sqrt2}\).
We'll make the following substitution, \[\large (x-1)=\frac{\sqrt3}{\sqrt2}\sec \theta\] So our denominator will do something like this, \[\large \left[2\left(\color{salmon}{(x-1)^2}-\dfrac{3}{2}\right)\right]^{3/2} \qquad \rightarrow \qquad \left[2\left(\color{salmon}{\frac{3}{2}\sec^2\theta}-\frac{3}{2}\right)\right]^{3/2}\]
\[\large \left[3\left(\sec^2\theta-1\right)\right]^{3/2} \qquad \rightarrow \qquad \left[3\left(\tan^2\theta\right)\right]^{3/2}\]Which simplifies to,\[\large \sqrt{27}\tan^3 \theta\]
Giving us an integral of,\[\large \frac{1}{\sqrt{27}}\int\limits\frac{dx}{\tan^3\theta}\]From here, we'll still need to deal with the dx.
\[\large (x-1)=\frac{\sqrt3}{\sqrt2}\sec \theta \qquad \rightarrow \qquad dx=\frac{\sqrt3}{\sqrt2}\sec\theta\tan\theta \; d \theta\] So our integral becomes,\[\large \frac{1}{\sqrt{27}}\cdot\frac{\sqrt3}{\sqrt2}\int\limits\limits\frac{\sec \theta \tan \theta \; d \theta}{\tan^3\theta}\]
So from here, cancel out a couple of tangents. Then convert everything to sines and cosines. At that point you'll have a couple options. You can do a nice easy U-sub, Or You can write your terms as \(\cot\theta \csc\theta\) which is a fairly easy integral.
There a bit more to be done to get through this. But I'm tired, and you're offline anyway XD lol.

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