# Thread: Help me with my calculus homework

1. ## Help me with my calculus homework

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..

2. 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.

3. 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??

4. 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?

5. 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??

6. 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
$

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

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

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..

7. 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?

8. 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
$y=\frac{\sqrt{t^2-5}}{\sqrt{11}}=
\frac{t-\sqrt{5}}{\sqrt{11}}$

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..
$\frac{\sqrt{11}}{11}\int{\frac{dt}{t}}$

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

am i correct??

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

10. 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??

11. 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.

12. 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??