$$\int \dfrac{1}{x^2\sqrt{x^2-5}}dx$$
Please, help

Question

loser66 · Answer

We have $tan^2t = sec^2 t-1$  and  $cot^2t = csc^2 t -1$

My question: why can't we use the second one to solve this problem?

inkyvoyd · Answer

attachment

loser66 · Answer

Thanks for the link @inkyvoyd . But when we use the second formula to solve it, we get a different answer. why?

inkyvoyd · Answer

What answer do you get?

loser66 · Answer

identity sin^2 t + cos^2 t =1 also.

jhonyy9 · Answer

like a first step rewrite 

x^2sqrt(x^2 -5) = sqrt(x^6 -5x4) 

than note sqrt(x^6 -5x^4) = u

loser66 · Answer

@jhonyy9  we go nowhere then.

loser66 · Answer

Ok, I show my work for the first formula

loser66 · Answer

Let $x =\sqrt5 sec t$ then $x^2 = 5sec^2 t$ and $\sqrt{x^2-5}=\sqrt{5(sec^2t -1)}=tant \sqrt5$

and then $dx= \sqrt5 sec t tant dt$

It becomes 
$$\int \dfrac{\sqrt5 sect t tant dt}{5sec^2t tant\sqrt5}=\dfrac{1}{5}\int cos tdt=\dfrac{1}{5}sin t+C$$

inkyvoyd · Answer

then x^2=25sec^2 t...

holsteremission · Answer

@Loser66 Nothing is wrong with using one formula over the other, but the second introduces a minus sign which could potentially lead to a mistake if you're not too careful.

loser66 · Answer

@HolsterEmission  I will show you the second formula

loser66 · Answer

Please, help me to check out the mistake

loser66 · Answer

The step to convert from sin t to x is easy. So that I don't waste your time on it. But the second formula leads me to cos x /5 
That is my problem

loser66 · Answer

Now, if let $x =\sqrt 5 csc t$, then $x^2 = 5 csc^2 t$ and $\sqrt{x^2 -5}= cot t\sqrt 5$

$dx=-\sqrt5 csc t cot t dt$ put them all

$$\int \dfrac{-\sqrt5 csct cot t dt}{5\sqrt5csc^2t cott}=-\dfrac{1}{5}\int \dfrac{1}{csc t}dt=\dfrac{-1}{5}\int sin tdt=\dfrac{1}{5}cost +C$$

loser66 · Answer

Where is my mistake?

agent0smith · Answer

You need to convert back to the original substitution... I'm willing to bet you'll see they're equivalent once you do that.

holsteremission · Answer

No mistake. With the second substitution, the last step would be to find $\cos\left(\mathrm{arccsc}\frac{x}{\sqrt5}\right)$, while the first substitution had you find $\sin\left(\mathrm{arcsec}\frac{x}{\sqrt5}\right)$.

loser66 · Answer

@agent0smith  what is the triangle you use to convert?

agent0smith · Answer

"The step to convert from sin t to x is easy. So that I don't waste your time on it"

This is your mistake. Assuming that they're wrong because you didn't complete it.

agent0smith · Answer

$$\large  \frac{ x }{ \sqrt 5 } = \csc t $$|dw:1476037705141:dw|

loser66 · Answer

|dw:1476037677052:dw|