Help me with my calculus homework

**magalingakosamath** · August 8th 2009, 06:13 PM

can you help me with this calculus problem?

here it is

integral of dy/sqrt[(11*y^2)+5]

i use algebraic substitution but i cant come up with an answer.. thnx in advance..

**Krizalid** · August 8th 2009, 06:24 PM

Okay, I'll help ya:

First off, put $y=\frac{\sqrt5}{\sqrt{11}}t,$ and the integrand (with no constants) will be $\frac1{\sqrt{t^2+1}}.$

Wonder how to integrate that? I know four methods, but the most common one is the straightforward trig. substitution.

**magalingakosamath** · August 8th 2009, 06:28 PM

oh.. i did not get that one.. can u explain the process.. how did you get off the constant??

i tried it with substitution but i just cant get it.. it will just re turn to the same equation but the sqrt is at the numerator..

ot:
how can u do the image something up there?? the image of the Y=sqrt of5/sqrt of 11 *t??

**Krizalid** · August 8th 2009, 06:33 PM

The imagination it's quite simple: put $y=\frac{\sqrt5}{\sqrt{11}}t$ (I'm not going to differentiate this, because I want you to try to see what I'm doing), and then $\frac{1}{\sqrt{11y^{2}+5}}=\frac{1}{\sqrt{11\left( \frac{\sqrt{5}}{\sqrt{11}}t \right)^{2}+5}}=\frac{1}{\sqrt{5t^{2}+5}}=\frac{1} {\sqrt{5}}\cdot \frac{1}{\sqrt{t^{2}+1}}.$ Does this make sense? Does it?

**magalingakosamath** · August 8th 2009, 06:40 PM

oh.. im sorry but where did you get the sqrt of 5/sqrt of 11*t

did you do something to it?? or you just let y be equal to that constant??

i mean i get the lat part cause its simple algebra but the first part??

**magalingakosamath** · August 8th 2009, 07:15 PM

ok this is easy to differetiate
$y=\frac{\sqrt5}{\sqrt{11}}t$

then after this one, the derivative of y

$dy=\frac{\sqrt5}{\sqrt{11}}$
am i right.. dont we need to substitute dy in the numerator since the original equation goes like this

$\frac{dy}{\sqrt{11y^{2}+5}}$

so im really wonderin were did you get
$y=\frac{\sqrt5}{\sqrt{11}}t<br />$

cant we let
$t={\sqrt{11y^2+5}}$

then it will become
$y=\frac{\sqrt{t^2-5}}{\sqrt{11}}=<br /> \frac{t-\sqrt{5}}{\sqrt{11}}$

then getting dy we will just use the formula integral of u/v is vdu-udv all over v^2 ayt?? that's the first method i did but i got confused.. then when you let y be equal to sqrt of 5 over 11 i also got confused..

**eXist** · August 8th 2009, 07:32 PM

Originally Posted by Krizalid

$y=\frac{\sqrt5}{\sqrt{11}}t,$

What he's doing here is straight forward. He's letting y = that expression, and then substituting it into the equation to make it easier to integrate. Looking at the new integral after substitution, can you integrate that?

**magalingakosamath** · August 8th 2009, 08:05 PM

yes i can, by trigonometric substitution..

ive already solved it but using algebraic substitution.. its simple.. but i dont know if its correct....

here is what i did.. from the original equation

i let
$t={\sqrt{11y^2+5}}$

then
$y=\frac{\sqrt{t^2-5}}{\sqrt{11}}=<br /> \frac{t-\sqrt{5}}{\sqrt{11}}$

i get the dy.. and it become
$dy=\frac{\sqrt{11}}{11}dt$

then i substitute the values of dy and t to the original equation and i get..
$\frac{\sqrt{11}}{11}\int{\frac{dt}{t}}$

then the answer will be

$\frac{\sqrt{11}}{11}\ln{t}+c$

am i correct??

**eXist** · August 8th 2009, 08:44 PM

Well take the derivative of your answer, do you get what you started with?

**magalingakosamath** · August 8th 2009, 09:04 PM

yes i get it.. i let the denomenator to be equal to "t" ,then i get the value of "y" in terms of "t".. then get the derivative of "y"...

so the original equation is
$\int{\frac{dy}{\sqrt{11y^{2}+5}}}$

i just substitute the value of dy and the denomenator so that i can integrate it easily.. then i came up with my answer.. so y do i still need to differentiate my answer??

**eXist** · August 8th 2009, 10:02 PM

I meant you can differentiate it to check to see if your answer is correct right? That's all I meant. You DO NOT have to differentiate your answer.

**magalingakosamath** · August 8th 2009, 10:47 PM

oh im so sorry bout that..

i get it., hehehe.. its the same.. but it has an excess of sqrt of 11.. maybe thats the constant c.. am i right??

Thread: Help me with my calculus homework

LinkBack

Thread Tools

Display

Help me with my calculus homework

Similar Math Help Forum Discussions

calculus homework

Calculus homework again...

Please help with calculus homework!!!!!

Calculus Homework Help Please

[SOLVED] AP Calculus Homework Help

Search Tags