Integral

**Mathstud28** · April 28th 2008, 12:09 PM

I have a portion of ad admittance exam coming up tomorrow and I was wondering if someone could give me some integrals I could solve and then you could check my work...I am getting fairly good at integrals but practice never hurt! Now I dont want ridculously difficult integrals...but I don't want too easy of ones either xD

If someoen could give me some I would be very greatful

Mathstud

**Moo** · April 28th 2008, 12:16 PM

Hello,

Try this one :

$\int_n^{n+1} \arctan \left(\frac{1}{\sqrt{t}} \right) dt$

There's nothing great about this, but take care of careless mistakes

**o_O** · April 28th 2008, 12:39 PM

A couple I found from the MIT integration bee from a while back:

$\int \left(\frac{1}{\ln x} + \ln \left(\ln x\right)\right)dx$

$\int \sin(101x) \cdot \sin^{99} (x) dx$

$\int e^{\arccos (x)} dx$

Have fun?

**Mathstud28** · April 28th 2008, 12:42 PM

Originally Posted by Moo

Hello,

Try this one :

$\int_n^{n+1} \arctan \left(\frac{1}{\sqrt{t}} \right) dt$

There's nothing great about this, but take care of careless mistakes

Thanks a lot Moo! I appreciate it...this is a fairly easy one...or at least I hope xD

$\int{arctan\bigg(\frac{1}{\sqrt{x}}\bigg)dx}=\int{ arcot(\sqrt{x})dx}$

Then you would go
Let $u=\sqrt{x}\Rightarrow{u^2=x}$
Then $dx=2udu$
So we have $\int{2u\cdot{arcot(u)du}}$

Using parts we have $\int{2u\cdot{arcot(u)}}du=u^2\cdot{arcot(u)}+\int\ frac{u^2}{1+u^2}$

THis is the same as
$u^2\cdot{arcot(u)}+\bigg[\int\frac{u^2+1}{u^2+1}du-\int\frac{1}{u^2+1}du\bigg]$

Which then obviously goes to

$u^2\cdot{arcot(u)}+u-arctan(u)$

Back subbing we get $x\cdot{arcot(\sqrt{x})}+\sqrt{x}-arctan(\sqrt{x})$

Whew I dont know why that took so long to write

...I showed all my steps just so you could critique me. Thanks

**Moo** · April 28th 2008, 12:50 PM

Actually, I've never used cot function...

Yes, the answer looks like yours

Try this one (I don't have the answer yet) :

$\int_0^1 \sqrt[3]{1-x^7}-\sqrt[7]{1-x^3} dx$

**Mathstud28** · April 28th 2008, 01:01 PM

Originally Posted by o_O

A couple I found from the MIT integration bee from a while back:

$\int \left(\frac{1}{\ln x} + \ln \left(\ln x\right)\right)dx$

$\int \sin(101x) \cdot \sin^{99} (x) dx$

$\int e^{\arccos (x)} dx$

Have fun?

Ok well lets do these one at a time...starting with number three

$\int{e^{arccos(x)}dx}$

Letting $u=arcos(x)\Rightarrow{cos(u)=x}$
Then $dx=-sin(u)$
So we have $\int-\sin(u)e^udu$
then by parts we haev $-sin(u)e^{u}+\int\cos(u)e^{u}du$
which then goes to $-sin(u)e^{u}+\cos(u)e^{u}-\int{-\sin(u)e^{u}du}$

Adding the like integrals and dividing by 2 we get $\int-\sin(u)e^{u}du=\frac{e^{u}}{2}(\cos(u)-\sin(u))$

back subbing we get $\int{e^{arccos(x)}dx}=\frac{e^{arccos(x)}}{2}(x-\sqrt{1-x^2})$

**Mathstud28** · April 28th 2008, 01:09 PM

Originally Posted by Moo

Actually, I've never used cot function...

Yes, the answer looks like yours

Try this one (I don't have the answer yet) :

$\int_0^1 \sqrt[3]{1-x^7}-\sqrt[7]{1-x^3} dx$

Aww moo...Do I really have to do that one...you would have to use the binomial coefficient power series

**Moo** · April 28th 2008, 01:12 PM

Originally Posted by Mathstud28

Aww moo...Do I really have to do that one...you would have to use the binomial coefficient power series

You said you wanted to train

I doubt your test will be about Krizalid-like integrals

**Mathstud28** · April 28th 2008, 01:16 PM

Originally Posted by Moo

You said you wanted to train

I doubt your test will be about Krizalid-like integrals

lasjflsdfkjasldfkjsadf;ljk......ok I will do it then

**Mathstud28** · April 28th 2008, 01:32 PM

Originally Posted by o_O

A couple I found from the MIT integration bee from a while back:

$\int \left(\frac{1}{\ln x} + \ln \left(\ln x\right)\right)dx$

$\int \sin(101x) \cdot \sin^{99} (x) dx$

$\int e^{\arccos (x)} dx$

Have fun?

For the third one we have $\int\frac{1}{\ln(x)}-\ln(\ln(x))dx$
Letting $u=\ln(x)\Rightarrow{e^u=x}$
then $dx=e^{u}$
so we have $\int{\frac{e^{u}}{u}+e^{u}\ln(u)du}$

which by splitting the integrals gives $\int{\frac{e^{u}}{u}du}+\int{e^{u}\ln(u)}dy$

leaving the first as it is we do integration by parts on the second to reveal

$\int{\frac{e^{u}}{u}+e^{u}\ln(u)}du=\int\frac{e^{u }}{u}du+\bigg[e^{u}\ln(u)-\int\frac{e^{u}}{u}du\bigg]$

cancelling like terms we get $\int{\frac{e^u}{u}+e^{u}\ln(u)}du=e^{u}\ln(u)$

Back subbing we get $\int\bigg[\frac{1}{\ln(x)}+\ln(\ln(x))dx\bigg]=e^{\ln(x)}\cdot{\ln(\ln(x))}=x\ln(\ln(x))+C$

**Mathstud28** · April 28th 2008, 02:11 PM

Originally Posted by Moo

Actually, I've never used cot function...

Yes, the answer looks like yours

Try this one (I don't have the answer yet) :

$\int_0^1 \sqrt[3]{1-x^7}-\sqrt[7]{1-x^3} dx$

$e\int_0^{1}\bigg[\ln\bigg(\frac{1}{x}\bigg)\bigg]^{\frac{1}{3}}dx-e\int_0^{1}\bigg[\ln\bigg(\frac{1}{x}\bigg)\bigg]^{\frac{1}{7}}dx$ ?

**Mathstud28** · April 28th 2008, 02:29 PM

Originally Posted by Mathstud28

Ok well lets do these one at a time...starting with number three

$\int{e^{arccos(x)}dx}$

Letting $u=arcos(x)\Rightarrow{cos(u)=x}$
Then $dx=-sin(u)$
So we have $\int-\sin(u)e^udu$
then by parts we haev $-sin(u)e^{u}+\int\cos(u)e^{u}du$
which then goes to $-sin(u)e^{u}+\cos(u)e^{u}-\int{-\sin(u)e^{u}du}$

Adding the like integrals and dividing by 2 we get $\int-\sin(u)e^{u}du=\frac{e^{u}}{2}(\cos(u)-\sin(u))$

back subbing we get $\int{e^{arccos(x)}dx}=\frac{e^{arccos(x)}}{2}(x-\sqrt{1-x^2})$

Originally Posted by Mathstud28

For the third one we have $\int\frac{1}{\ln(x)}-\ln(\ln(x))dx$
Letting $u=\ln(x)\Rightarrow{e^u=x}$
then $dx=e^{u}$
so we have $\int{\frac{e^{u}}{u}+e^{u}\ln(u)du}$

which by splitting the integrals gives $\int{\frac{e^{u}}{u}du}+\int{e^{u}\ln(u)}dy$

leaving the first as it is we do integration by parts on the second to reveal

$\int{\frac{e^{u}}{u}+e^{u}\ln(u)}du=\int\frac{e^{u }}{u}du+\bigg[e^{u}\ln(u)-\int\frac{e^{u}}{u}du\bigg]$

cancelling like terms we get $\int{\frac{e^u}{u}+e^{u}\ln(u)}du=e^{u}\ln(u)$

Back subbing we get $\int\bigg[\frac{1}{\ln(x)}+\ln(\ln(x))dx\bigg]=e^{\ln(x)}\cdot{\ln(\ln(x))}=x\ln(\ln(x))+C$

Are these right o_O?

**o_O** · April 28th 2008, 03:04 PM

Looks good to me. You made some typos but that's probably due to messing up while typing up your work.

How about this one:
$\int_{0}^{\infty} xe^{-x^{3}}dx$

**Mathstud28** · April 28th 2008, 03:13 PM

Originally Posted by o_O

Looks good to me. You made some typos but that's probably due to messing up while typing up your work.

How about this one:
$\int_{0}^{\infty} xe^{-x^{3}}dx$

$\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{3n+2}}{(3n+2)n!}\bigg|_{0}^{\infty}$

Is that right?

**Krizalid** · April 28th 2008, 03:45 PM

Do you know when you have to apply power series for an integral? Did you think first to make a substitution?

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