# Iterated integral 2

#### Petrus

##### Well-known member
Hello MHB,
$$\displaystyle \int_0^1\int_0^1 \frac{xy}{\sqrt{x^2+y^2+1}} dxdy$$
I start with subsitate $$\displaystyle u=x <=> du=dx$$ and $$\displaystyle du= \frac{y}{\sqrt{x^2+y^2+1}} <=>u=y\ln\sqrt{x^2+y^2+1}$$ so we got integrate by part that
$$\displaystyle xy\ln\sqrt{x^2+y^2+1}]_0^1-\int_0^1\frac{y}{\sqrt{x^2+y^2+1}}dx$$
and we got
$$\displaystyle [xy\ln\sqrt{x^2+y^2+1}]_0^1-[y\ln{\sqrt{x^2+y^2+1}}]_0^1$$
Remember that we solve dx. Is this correct?

Regards,

#### MarkFL

Staff member
Hello Petrus,

I think I would first write the integral as:

$$\displaystyle \int_0^1 y\left(\int_0^1\frac{x}{\sqrt{x^2+y^2+1}}\,dx \right)\,dy$$

Now, on the inner integral, consider the substitution:

$$\displaystyle u=x^2+y^2+1$$

What do you have now?

#### Sudharaka

##### Well-known member
MHB Math Helper
Hello MHB,
$$\displaystyle \int_0^1\int_0^1 \frac{xy}{\sqrt{x^2+y^2+1}} dxdy$$
I start with subsitate $$\displaystyle u=x <=> du=dx$$ and $$\displaystyle du= \frac{y}{\sqrt{x^2+y^2+1}} <=>u=y\ln\sqrt{x^2+y^2+1}$$ so we got integrate by part that
$$\displaystyle xy\ln\sqrt{x^2+y^2+1}]_0^1-\int_0^1\frac{y}{\sqrt{x^2+y^2+1}}dx$$
and we got
$$\displaystyle [xy\ln\sqrt{x^2+y^2+1}]_0^1-[y\ln{\sqrt{x^2+y^2+1}}]_0^1$$
Remember that we solve dx. Is this correct?

Regards,
Hi Petrus,

Substituting $$x=u$$ won't give you anything useful since you are just replacing $$x$$ by $$u$$. You can try solving this problem with the substitution Mark has given, but let me suggest a slightly different method.

Differentiate $$\sqrt{x^2+y^2+1}$$ and see what you get and try to use that result in solving the integral.

#### Petrus

##### Well-known member
Hello Petrus,

I think I would first write the integral as:

$$\displaystyle \int_0^1 y\left(\int_0^1\frac{x}{\sqrt{x^2+y^2+1}}\,dx \right)\,dy$$

Now, on the inner integral, consider the substitution:

$$\displaystyle u=x^2+y^2+1$$

What do you have now?
Hmm... $$\displaystyle u=x^2+y^2+1$$ $$\displaystyle du =2x$$
$$\displaystyle \int_0^1 \frac{y}{2} \int_0^1 \frac{1}{\sqrt{u}}du$$
hmmm is this correct?

Regards,

Last edited:

#### Petrus

##### Well-known member
Hi Petrus,

Substituting $$x=u$$ won't give you anything useful since you are just replacing $$x$$ by $$u$$. You can try solving this problem with the substitution Mark has given, but let me suggest a slightly different method.

Differentiate $$\sqrt{x^2+y^2+1}$$ and see what you get and try to use that result in solving the integral.
Hello Sudharaka,
I can't see how you can see this will give no progress. Have I miss something with 'integrate by part', I honestly doubt what method I will use when I come to this kind of question. Function divide by function or function multiplicate by function ( When you integrate)

Regards,

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Be careful to the boundaries of integration when you make a substitution.

#### Petrus

##### Well-known member
Be careful to the boundaries of integration when you make a substitution.
Yeah I am aware of that I will just subsitute back after I antiderivate, that's why I did not rewrite the limit of integrate

Regards,

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Hi Petrus,

Substituting $$x=u$$ won't give you anything useful since you are just replacing $$x$$ by $$u$$.

I think what Petrus is trying to do is integration by parts .

If you differentiate $$\displaystyle y \sqrt {x^2+y^2+1}$$ w.r.t to $x$ what do we get ?

#### Petrus

##### Well-known member
So far I got:
$$\displaystyle u=x^2+y^2+1$$ $$\displaystyle du =2x$$
$$\displaystyle \int_0^1 \frac{y}{2} \int_0^1 \frac{1}{\sqrt{u}}du$$
$$\displaystyle \int_0^1 \frac{y}{2} [2\sqrt{u}]$$ If we subsitute it back we get:
$$\displaystyle \int_0^1 \frac{y}{2} [2\sqrt{x^2+y^2+1}]_0^1$$
now we got:
$$\displaystyle \int_0^1y\sqrt{2+y^2}-y\sqrt{1+y^2}$$
We subsitute $$\displaystyle u=2+y^2$$ $$\displaystyle du=2y$$
so we got
$$\displaystyle \frac{1}{2}\int_0^1 \sqrt{u}-\sqrt{u-1}$$
If I antiderivate that I get
$$\displaystyle \frac{1}{2}[\frac{2u^{1.5}}{3}-\frac{2(u-1)^{1.5}}{3}]$$
and when i subsitute back and put the limits I get wrong answer, have I done something wrong here?

Regards,

Last edited:

#### Sudharaka

##### Well-known member
MHB Math Helper
Hello Sudharaka,
I can't see how you can see this will give no progress. Have I miss something with 'integrate by part', I honestly doubt what method I will use when I come to this kind of question. Function divide by function or function multiplicate by function ( When you integrate)

Regards,
Ah, you are trying to integrate by parts, that makes sense. I think I misunderstood your attempt to integrate by parts when reading the line,

........I start with subsitate $$\displaystyle u=x <=> du=dx$$ and......
in your first post. Sorry about that. What I was suggesting is to observe that,

$\frac{d}{dx}\sqrt{x^2+a}=\frac{x}{\sqrt{x^2+a}}$

where $$a$$ is a constant. That is,

$\int \frac{x}{\sqrt{x^2+a}}\,dx=\sqrt{x^2+a}+C$

where $$C$$ is an arbitrary constant. I thought that this is something which isn't hard to observe. Using this you can solve the inner integral of your iterated integral.

#### Petrus

##### Well-known member
Thanks Sudharaka,
I did never think about that! Can someone control my post #9 what I have done wrong?

Regards,

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Thanks Sudharaka,
I did never think about that! Can someone control my post #9 what I have done wrong?

Regards,
I dont see any mistakes .

#### Petrus

##### Well-known member
I dont see any mistakes .
I looked at wrong answer in facit haha.... Thanks got it now I will try solve it with integrate by part now