baldymcgee6
  • baldymcgee6
Use an inverse trig substitution on the integral...
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
baldymcgee6
  • baldymcgee6
|dw:1358909488344:dw|
abb0t
  • abb0t
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.
baldymcgee6
  • baldymcgee6
complete the square in the denominator... okkkk

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

baldymcgee6
  • baldymcgee6
you're going to have to refresh my memory on how to do that :/
baldymcgee6
  • baldymcgee6
|dw:1358910770499:dw|
abb0t
  • abb0t
\[ax^2+bx+c \] \[(\frac{ b }{ 2 })^2 \]
baldymcgee6
  • baldymcgee6
i'm sorry, i forget what to do...
abb0t
  • abb0t
divide by two and square the answer of the 2nd term (b).
abb0t
  • abb0t
then add it to C.
baldymcgee6
  • baldymcgee6
|dw:1358910964480:dw|
abb0t
  • abb0t
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.
abb0t
  • abb0t
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.
baldymcgee6
  • baldymcgee6
|dw:1358914184400:dw|@abb0t
abb0t
  • abb0t
Now use u-substitution.
baldymcgee6
  • baldymcgee6
of what?
baldymcgee6
  • baldymcgee6
u - x-1 ?
abb0t
  • abb0t
yeah
baldymcgee6
  • baldymcgee6
u = x-1 ***
baldymcgee6
  • baldymcgee6
i don't see where that takes us.
baldymcgee6
  • baldymcgee6
|dw:1358914405152:dw|
baldymcgee6
  • baldymcgee6
@zepdrix ?
zepdrix
  • zepdrix
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.\[\large \int\limits\frac{dx}{\left[2(x-1)^2-3\right]^{3/2}}\]
zepdrix
  • zepdrix
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.\[\large \int\limits\limits\frac{dx}{\left[2\left((x-1)^2-\dfrac{3}{2}\right)\right]^{3/2}}\]
zepdrix
  • zepdrix
"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 \(\dfrac{\sqrt3}{\sqrt2}\).
zepdrix
  • zepdrix
We'll make the following substitution, \[\large (x-1)=\frac{\sqrt3}{\sqrt2}\sec \theta\] So our denominator will do something like this, \[\large \left[2\left(\color{salmon}{(x-1)^2}-\dfrac{3}{2}\right)\right]^{3/2} \qquad \rightarrow \qquad \left[2\left(\color{salmon}{\frac{3}{2}\sec^2\theta}-\frac{3}{2}\right)\right]^{3/2}\]
zepdrix
  • zepdrix
\[\large \left[3\left(\sec^2\theta-1\right)\right]^{3/2} \qquad \rightarrow \qquad \left[3\left(\tan^2\theta\right)\right]^{3/2}\]Which simplifies to,\[\large \sqrt{27}\tan^3 \theta\]
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
\[\large (x-1)=\frac{\sqrt3}{\sqrt2}\sec \theta \qquad \rightarrow \qquad dx=\frac{\sqrt3}{\sqrt2}\sec\theta\tan\theta \; d \theta\] So our integral becomes,\[\large \frac{1}{\sqrt{27}}\cdot\frac{\sqrt3}{\sqrt2}\int\limits\limits\frac{\sec \theta \tan \theta \; d \theta}{\tan^3\theta}\]
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
There a bit more to be done to get through this. But I'm tired, and you're offline anyway XD lol.

Looking for something else?

Not the answer you are looking for? Search for more explanations.