# Thread: Integral

1. ## Integral

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

2. 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

3. 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?

4. 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

5. Actually, I've never used cot function...
Yes, the answer looks like yours

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

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

6. 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 $\int{e^{arccos(x)}dx}=\frac{e^{arccos(x)}}{2}(x-\sqrt{1-x^2})$

7. 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

8. 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

9. 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

10. 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 $\int\bigg[\frac{1}{\ln(x)}+\ln(\ln(x))dx\bigg]=e^{\ln(x)}\cdot{\ln(\ln(x))}=x\ln(\ln(x))+C$

11. 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) :

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

12. 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?

13. 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:
$\int_{0}^{\infty} xe^{-x^{3}}dx$

14. 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?

15. 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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