# Simple Integral question w/ parts

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• Sep 2nd 2008, 11:10 PM
matt3D
Simple Integral question w/ parts
Okay, I have a question. Is $\int arctan{5y}$ equal to $\frac{1}{1+5y^2}\times10y$?
I am working on this problem,
$\int x^2arctan{2x}dx$, it's actually tan^-1(2x) but I'm not sure how to insert the negative to make it the inverse.
I made $u=2y,\ du=2dy,\ dv=arctan{5y},\ and\ v=\frac{10y}{a+5y^2}$

-Matt
• Sep 2nd 2008, 11:33 PM
flyingsquirrel
Hello,
Quote:

Originally Posted by matt3D
Okay, I have a question. Is $\int arctan{5y}$ equal to $\frac{1}{1+5y^2}\times10y$?

No, we have $\frac{\mathrm{d}}{\mathrm{d}y}\left[\arctan (5y)\right]=\frac{5}{1+25y^2}$ or $\arctan (5y) = \int_0^{5y}\frac{1}{1+t^2}\,\mathrm{d}t$.
Quote:

I am working on this problem,
$\int x^2arctan{2x}dx$, it's actually tan^-1(2x) but I'm not sure how to insert the negative to make it the inverse.
Write $$\tan^{-1}$$ or $$\arctan$$ .
Quote:

I made $u=2y,\ du=2dy,\ dv=arctan{5y},\ and\ v=\frac{10y}{a+5y^2}$
Am I heading in the right direction?
I don't think so since you don't know an antiderivative of $\arctan$. One possibility is to see that if you let $v=\arctan (2x)$ and $u'=x^2$ then $v'=\frac{2}{1+4x^2}$ and $u=\frac{x^3}{3}$ so $uv'$ is a rational fraction : we don't have to integrate $\arctan$ anymore.

Does it help ?
• Sep 2nd 2008, 11:50 PM
Prove It
Quote:

Originally Posted by matt3D
Okay, I have a question. Is $\int arctan{5y}$ equal to $\frac{1}{1+5y^2}\times10y$?
I am working on this problem,
$\int x^2arctan{2x}dx$, it's actually tan^-1(2x) but I'm not sure how to insert the negative to make it the inverse.
I made $u=2y,\ du=2dy,\ dv=arctan{5y},\ and\ v=\frac{10y}{a+5y^2}$

-Matt

According to my book of integrals, an integral of $\arctan x$...

$\int \arctan x\, dx = x\arctan x - \frac{1}{2} \log (1+x^2)$.

Does that help?
• Sep 3rd 2008, 07:39 PM
matt3D
Oops, I mixed my problem up. This is what I am currently trying to figure out: $\int2ytan^{-1}{5y}dy$
• Sep 3rd 2008, 08:00 PM
matt3D
Okay so, $u=\ 2y,\ du=\ 2dy,\ dv=\ tan^{-1}{5y},\ and\ v=5ytan^{-1}{5y}-\frac {1}{2}\ln{(1+25y^4)}+c$ correct? This is looking more difficult then I thought! So now: $2y\times 5ytan^{-1}{5y}-\frac {1}{2}\ln{(1+25y^4)}-\frac {1}{2}\int 5ytan^{-1}{5y}-\frac {1}{2}\ln{(1+25y^4)}dy$ I put the $\frac {1}{2}$ in front of the integral to get rid of the 2 from du which is 2dy...
• Sep 3rd 2008, 08:22 PM
Chris L T521
Quote:

Originally Posted by matt3D
Okay so, $u=\ 2y,\ du=\ 2dy,\ dv=\ tan^{-1}{5y},\ and\ v=5ytan^{-1}{5y}-\frac {1}{2}\ln{(1+25y^{\color{red}4})}+c$ correct? This is looking more difficult then I thought! So now: $2y\times 5ytan^{-1}{5y}-\frac {1}{2}\ln{(1+25y^{\color{red}4})}-\frac {1}{2}\int 5ytan^{-1}{5y}-\frac {1}{2}\ln{(1+25y^{\color{red}4})}dy$ I put the $\frac {1}{2}$ in front of the integral to get rid of the 2 from du which is 2dy...

The terms in red should be 2, not 4.

Now, from here, do the following:

We found that $\int 5y\tan^{-1}(5y)\,dy=$ $10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}-\frac {1}{2}{\color{red}\int} \left({\color{red}5y\tan^{-1}(5y)}-\tfrac {1}{2}\ln{(1+25y^2)}\right){\color{red}\,dy}$

The integral reappears on the right side! So get it on the left side!

$\implies \tfrac{3}{2}\int 5y\tan^{-1}(5y)\,dy=10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}+\tfrac{1}{4}{\color{red}\int\ ln{(1+25y^2)}\,dy}$

Before we get our answer, try to evaluate the integral I highlighted in red.

Hint : apply integration by parts! Let $u=\ln(1+25y^2)$.....
• Sep 3rd 2008, 08:32 PM
o_O
$\int 2y \tan^{-1} (5y) \ dy$

First, let's make the sub: $\begin{array}{lll} s = 5y & \Rightarrow & ds = 5 dy \iff dy = \displaystyle \frac{ds}{5} \\ y = \displaystyle \frac{s}{5} \end{array}$

and we get: $\int 2 \left(\frac{s}{5}\right) \tan^{-1} s \frac{ds}{5} \ \ = \ \ \frac{1}{25} \int 2s \tan^{-1}s \ ds$

Now you can use integration by parts: $\begin{array}{ll} u = \tan^{-1} s & dv = 2s ds \\ du = \displaystyle \frac{1}{1+s^2} \ ds & v = s^2 \end{array}$
• Sep 3rd 2008, 08:40 PM
matt3D
Quote:

Originally Posted by Chop Suey
Bad news: tan isn't a variable.

The $5y\tan^{-1}{5y}$ is an antiderivative of $\tan^{-1}{5y}$. Try differentiating and see for yourself.

You have to be wise when you choose what u and dv should be. You're integrating $\tan^{-1}{5y}$, but you don't know its integral, so how could you think of it as dv?!

Yea, I see what you mean! (Nod)
Quote:

Originally Posted by Chris L T521
The terms in red should be 2, not 4.

Now, from here, do the following:

We found that $\int 5y\tan^{-1}(5y)\,dy=$ $10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}-\frac {1}{2}{\color{red}\int} \left({\color{red}5y\tan^{-1}(5y)}-\tfrac {1}{2}\ln{(1+25y^2)}\right){\color{red}\,dy}$

The integral reappears on the right side! So get it on the left side!

$\implies \tfrac{3}{2}\int 5y\tan^{-1}(5y)\,dy=10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}+\tfrac{1}{4}{\color{red}\int\ ln{(1+25y^2)}\,dy}$

Before we get our answer, try to evaluate the integral I highlighted in red.

Hint : apply integration by parts! Let $u=\ln(1+25y^2)$.....

How does it become 3/2 on the left side? I know its because you put the integral to the left though like you said, but how exactly?
Quote:

Originally Posted by o_O
$\int 2y \tan^{-1} (5y) \ dy$

First, let's make the sub: $\begin{array}{lll} s = 5y & \Rightarrow & ds = 5 dy \iff dy = \displaystyle \frac{ds}{5} \\ y = \displaystyle \frac{s}{5} \end{array}$

and we get: $\int 2 \left(\frac{s}{5}\right) \tan^{-1} s \frac{ds}{5} \ \ = \ \ \frac{1}{25} \int 2s \tan^{-1}s \ ds$

Now you can use integration by parts: $\begin{array}{ll} u = \tan^{-1} s & dv = 2s ds \\ du = \displaystyle \frac{1}{1+s^2} \ ds & v = s^2 \end{array}$

Okay, I never thought of it like that, I'm going to try that now.
• Sep 3rd 2008, 08:46 PM
Chris L T521
Quote:

Originally Posted by Chris L T521
The terms in red should be 2, not 4.

Now, from here, do the following:

We found that $\int 5y\tan^{-1}(5y)\,dy=$ $10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}{\color{red}-\tfrac{1}{2}\int} \left({\color{red}5y\tan^{-1}(5y)}-\tfrac {1}{2}\ln{(1+25y^2)}\right){\color{red}\,dy}$

The integral reappears on the right side! So get it on the left side!

$\implies \tfrac{3}{2}\int 5y\tan^{-1}(5y)\,dy=10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}+\tfrac{1}{4}{\color{red}\int\ ln{(1+25y^2)}\,dy}$

Before we get our answer, try to evaluate the integral I highlighted in red.

Hint : apply integration by parts! Let $u=\ln(1+25y^2)$.....

Quote:

Originally Posted by matt3D

How does it become 3/2 on the left side? I know its because you put the integral to the left though like you said, but how exactly?

We see that we have the term $-\tfrac{1}{2}\int (5y\tan^{-1}(5y))\,dy$ on the right side of the equation and $\int 5y\tan^{-1}(5y)\,dy$ on the left side.

Add $\tfrac{1}{2}\int (5y\tan^{-1}(5y))\,dy$ to both sides. The integral will disappear on the right and on the left side of the equation we get $\int 5y\tan^{-1}(5y)\,dy+\tfrac{1}{2}\int (5y\tan^{-1}(5y))\,dy=\tfrac{3}{2}\int (5y\tan^{-1}(5y))\,dy$

I hope this makes sense.

--Chris
• Sep 3rd 2008, 08:50 PM
matt3D
Quote:

Originally Posted by Chris L T521
We see that we have the term $-\tfrac{1}{2}\int (5y\tan^{-1}(5y))\,dy$ on the right side of the equation and $\int 5y\tan^{-1}(5y)\,dy$ on the left side.

Add $\tfrac{1}{2}\int (5y\tan^{-1}(5y))\,dy$ to both sides. The integral will disappear on the right and on the left side of the equation we get $\int 5y\tan^{-1}(5y)\,dy+\tfrac{1}{2}\int (5y\tan^{-1}(5y))\,dy=\tfrac{3}{2}\int (5y\tan^{-1}(5y))\,dy$

I hope this makes sense.

--Chris

Ah, I see. Thank you. I should have just remembered the 2/2 on the left to add to 1/2
• Sep 3rd 2008, 08:58 PM
matt3D
Quote:

Originally Posted by o_O
$\int 2y \tan^{-1} (5y) \ dy$

First, let's make the sub: $\begin{array}{lll} s = 5y & \Rightarrow & ds = 5 dy \iff dy = \displaystyle \frac{ds}{5} \\ y = \displaystyle \frac{s}{5} \end{array}$

and we get: $\int 2 \left(\frac{s}{5}\right) \tan^{-1} s \frac{ds}{5} \ \ = \ \ \frac{1}{25} \int 2s \tan^{-1}s \ ds$

Now you can use integration by parts: $\begin{array}{ll} u = \tan^{-1} s & dv = 2s ds \\ du = \displaystyle \frac{1}{1+s^2} \ ds & v = s^2 \end{array}$

So, I get $s^2(tan^{-1}s)-\int \frac {s^2}{1+s^2}ds$ but I need some help integrating the integral, it looks very similar to $tan^{-1}{s}$ except with the $s^2$ on the numerator.
• Sep 3rd 2008, 09:03 PM
Chris L T521
Quote:

Originally Posted by matt3D
So, I get $s^2(tan^{-1}s)-\int \frac {s^2}{1+s^2}ds$ but I need some help integrating the integral, it looks very similar to $tan^{-1}{s}$ except with the $s^2$ on the numerator.

One way to deal with the integral would be to apply long division to the integrand:

I leave it for you to verify that $\frac{s^2}{1+s^2}=1-\frac{1}{1+s^2}$

Now integrate $\int\left(1-\frac{1}{1+s^2}\right)\,ds$. It should be a piece of cake now.

I hope this makes sense! (Sun)

--Chris
• Sep 3rd 2008, 09:07 PM
11rdc11
Try $\int{ds} - \int\frac{ds}{s^2+1} = \int\frac{s^2}{1+s^2}$

oops I was too slow
• Sep 3rd 2008, 09:24 PM
matt3D
Alright, so I finally get: $\frac {25y^2(tan^{-1}{5y})+tan^{-1}{5y}-5y}{25}$ Now I'm a little confused about the division though, 11rdc11, could you explain your last post a little more?
Thank you guys for all of your help! I really do appreciate it.
• Sep 3rd 2008, 09:28 PM
Chris L T521
Quote:

Originally Posted by matt3D
Alright, so I finally get: $\frac {25y^2(tan^{-1}{5y})+tan^{-1}{5y}-5y}{25}+{\color{red}C}$ Now I'm a little confused about the division though, 11rdc11, could you explain your last post a little more?
Thank you guys for all of your help! I really do appreciate it.

(Yes)

Where are you stuck in the division process?

--Chris
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