Evaluating Integral using Parts

[doublehh06]

Pardon my use of the program! I am new to Physics Forums!

Homework Statement

EVALUATE

The Integral of: Square root of (x^2 + 2x)
The Integral of: x * Square root of (x^2 + 7)

Homework Equations

Integrating by Parts Method

The Integral of udv = u*v - the integral of v*du

The Attempt at a Solution

I am confused as to which parts should be u and which should be dv. It is just with square roots when this happens and I just would like some guidance!

 

[micromass]

There's no need to integrate by parts. Just use trigonometric substitution. For 1, you will need to complete the square first...

 

[FLUndergrad]

for x(sqrt[x^2+7]) you can also use simple substitution where u=(x^2+7) and du=2x, which makes it into the integral of (1/2)(u)^(1/2)du, which gives you (1/3)(x^2+7)^(3/2)

 

[doublehh06]

Micromass -

When you say complete the square, what do you mean?

If I complete the square, that would give me the square root of (x+1)^2 - 1, right?
Then do I use trigonometic substitution?

 

[micromass]

Micromass -

When you say complete the square, what do you mean?

If I complete the square, that would give me the square root of (x+1)^2 - 1, right?
Then do I use trigonometic substitution?

Right, but before you do trigonometric substitution, first substitute u=x+1...

 

[doublehh06]

I have the answer (found in my textbook for the homework problem) which is:

1/2 * (x+1) * The Integral of: Square Root of (x^2 + 2x) - 1/2 the natural log of (x +1 + square root of (x^2 + 2x)) + C

Sorry... I have tried to do this even with the u substitution. I am just completely clueless with how to get to this point!

 

[micromass]

With the u-substitution, you've got

$$\int{\sqrt{u^2-1}du}$$

Now, try a trigonometric substitution. Something like u=sec(v)...

 

[doublehh06]

So then with the U-Substitution,

We would have sec²u-1du beneath the square root. Does this then turn into tan²udu beneath the square root and then I perform the integral or try to do substitution again, etc.?

 

[doublehh06]

The Final Answer is:

\displaystyle{\frac{1}{2}}(x+1)\sqrt{x^2+2x}-\displaystyle{\frac{1}{2}}\ln{|x+1+\sqrt{x^2+2x}|}+C

This is in the form of

\int{udv}=uv-\int{vdu}

Which is valid for integration by parts.. I know we have to do trigonometric substitution also, but I don't know how we will get to the final answer using trig substitution.

I did all of the codes using the PF LaTeX Preview, but they still aren't showing up right here in the forum. Sorry!

 

[micromass]

So then with the U-Substitution,

We would have sec²u-1du beneath the square root. Does this then turn into tan²udu beneath the square root and then I perform the integral?

Yes, just substitute u=sec(v) and perform the integral...

 

[Saladsamurai]

The Final Answer is:

$$\displaystyle{\frac{1}{2}}(x+1)\sqrt{x^2+2x}-\displaystyle{\frac{1}{2}}\ln{|x+1+\sqrt{x^2+2x}|}+C$$

This is in the form of

$$\int{udv}=uv-\int{vdu}$$

Which is valid for integration by parts.. I know we have to do trigonometric substitution also, but I don't know how we will get to the final answer using trig substitution.

I did all of the codes using the PF LaTeX Preview, but they still aren't showing up right here in the forum. Sorry!

By adding the LaTeX tags you can get this to display properly. Click on the LaTeX image (or just quote me) on your quote above to see the code that generated each image

 

[Liquidxlax]

So then with the U-Substitution,

We would have sec²u-1du beneath the square root. Does this then turn into tan²udu beneath the square root and then I perform the integral or try to do substitution again, etc.?

$$\sqrt{tan^{2}u}du=tanudu$$

what is the integral of tanudu? and re substitute

 

[doublehh06]

The integral of tanudu is -ln|cos{u}|+C which then turns into -ln|cos{x+1}|+C.

This still isn't the right answer which can be seen in the previous quote by saladsamurai.
Something is still going wrong!

 

[Bohrok]

If you use a u-substitution like u = sec v, you must find du from that and replace the du from the previous integral you had.

 

[doublehh06]

I'm completely lost still. I can't get past the answer I had before. I just don't understand how to do this combination of integration by parts AND trigonometric substitution.

 

[micromass]

You did your substitution u=sec(v) wrong. You need to replace the du to...
Can you give me the general formula for substitution??

 

[doublehh06]

I don't know what the general formula is... I just know that if you substitute in for u than your du value changes also. I have done substitution before and been fine, but I can't give you a general formula.

 

[micromass]

Fine, if you substitute u=sec(v). You shouldn't forget to change the du... If u=sec(v), then what does du equal?

 

[doublehh06]

du = sec(v)tan(v)

 

[micromass]

Yes, so your integral becomes

$$\int{\tan^2(v)\sec(v)dv}$$

Try to solve this integral...

 

[doublehh06]

Do I have to perform yet another substitution to solve this integral? substituting a variable in for sec (v) like t making the new integral:

t^2 dt?

 

[micromass]

Start by writing everything in cos(v). Thus change that tan(v) and sec(v), such that you arrive with something with only cos(v).

 

[doublehh06]

My steps to get something with only cos(v) are as follows:

sin^2(v)/cos^2(v) * (1/cos(v))dv

(1-cos^2(v))/cos^2(v) * (1/cos(v))dv

(1-cos^2(v))/cos^3(v) dv

1/cos^3(v) - 1/cos(v)

Is this the right integral I should end up with?

 

[micromass]

Yes, that would be correct!

 

[doublehh06]

Now, once I perform that integral do I just resubstitute a lot to get to the correct answer? I mean, when finding the antiderivative of something which has a trig function in it, you almost always get something with a trig function still in it. I am still not seeing how this will lead to the final answer which has no trig function in it at all.

 

[doublehh06]

I've got some work done on my own now and have started to resubstitute to get back to having the function in terms of x.

Can anyone tell me what tan * sec^-1(x) is equal to?

 

[Mark44]

I've got some work done on my own now and have started to resubstitute to get back to having the function in terms of x.

Can anyone tell me what tan * sec^-1(x) is equal to?

As it stands, it doesn't equal anything. You can't have tan all by itself any more than you can have $\sqrt$ all by itself. tan is a function, so you have to specify what it's operating on.

 

[Mark44]

du = sec(v)tan(v)

That would be du = sec(v)tan(v)dv
Don't leave this off! If you do, it will come around and bite you later.

 

[doublehh06]

I figured it out on my own! Thanks to everyone who helped on this problem! :)