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..
Okay, I'll help ya:
First off, put $\displaystyle y=\frac{\sqrt5}{\sqrt{11}}t,$ and the integrand (with no constants) will be $\displaystyle \frac1{\sqrt{t^2+1}}.$
Wonder how to integrate that? I know four methods, but the most common one is the straightforward trig. substitution.
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??
The imagination it's quite simple: put $\displaystyle y=\frac{\sqrt5}{\sqrt{11}}t$ (I'm not going to differentiate this, because I want you to try to see what I'm doing), and then $\displaystyle \frac{1}{\sqrt{11y^{2}+5}}=\frac{1}{\sqrt{11\left( \frac{\sqrt{5}}{\sqrt{11}}t \right)^{2}+5}}=\frac{1}{\sqrt{5t^{2}+5}}=\frac{1} {\sqrt{5}}\cdot \frac{1}{\sqrt{t^{2}+1}}.$ Does this make sense? Does it?
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??
ok this is easy to differetiate
$\displaystyle y=\frac{\sqrt5}{\sqrt{11}}t$
then after this one, the derivative of y
$\displaystyle dy=\frac{\sqrt5}{\sqrt{11}}$
am i right.. dont we need to substitute dy in the numerator since the original equation goes like this
$\displaystyle \frac{dy}{\sqrt{11y^{2}+5}}$
so im really wonderin were did you get
$\displaystyle y=\frac{\sqrt5}{\sqrt{11}}t
$
cant we let
$\displaystyle t={\sqrt{11y^2+5}}$
then it will become
$\displaystyle y=\frac{\sqrt{t^2-5}}{\sqrt{11}}=
\frac{t-\sqrt{5}}{\sqrt{11}}$
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..
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
$\displaystyle t={\sqrt{11y^2+5}}$
then
$\displaystyle y=\frac{\sqrt{t^2-5}}{\sqrt{11}}=
\frac{t-\sqrt{5}}{\sqrt{11}}$
i get the dy.. and it become
$\displaystyle dy=\frac{\sqrt{11}}{11}dt$
then i substitute the values of dy and t to the original equation and i get..
$\displaystyle \frac{\sqrt{11}}{11}\int{\frac{dt}{t}}$
then the answer will be
$\displaystyle \frac{\sqrt{11}}{11}\ln{t}+c$
am i correct??
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
$\displaystyle \int{\frac{dy}{\sqrt{11y^{2}+5}}}$
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??