Why my result is not correct?

[Vali]

I have the function f-infinity, -1) -> R $\displaystyle f(x)=\frac{1}{x\sqrt{x^{2}-1}}$
I have to find the primitives of this function.So I need to calculate $\displaystyle \int \frac{1}{x\sqrt{x^{2}-1}}$
I solved it with u=sqrt(x^2 - 1) and I got $\displaystyle arctan\sqrt{x^{2}-1} + C$ but the right answer is arcsin(1/x) + C, why?
I tried also to solve it using t=1/x but I didn't get too far.

 

[MarkFL]

If we use:

[MATH]u=\sqrt{x^2-1}[/MATH]
then:

[MATH]du=\frac{x}{\sqrt{x^2-1}\,dx}[/MATH]
And the integral becomes:

[MATH]I=\int \frac{1}{u^2+1}\,du=\arctan(u)+C=\arctan(\sqrt{x^2-1})+C[/MATH]
So, I agree with your result. Could it be that the difference between your result and what you call the right answer is just a constant?

 

[Vali]

The original exercise is written like this, and the right anwer is B

 

[MarkFL]

It appears to me that you and I assumed $1<x$, and so we want:

[MATH]-\int \frac{1}{x\sqrt{x^2-1}}\,dx=-\arctan(\sqrt{x^2-1})+C[/MATH]
And then we could observe that:

[MATH]\arctan\left(\frac{1}{\sqrt{x^2-1}}\right)=\frac{\pi}{2}-\arctan(\sqrt{x^2-1})[/MATH]
Hence:

[MATH]-\int \frac{1}{x\sqrt{x^2-1}}\,dx=\arctan\left(\frac{1}{\sqrt{x^2-1}}\right)+C[/MATH]
And this is equivalent to:

[MATH]-\int \frac{1}{x\sqrt{x^2-1}}\,dx=\arcsin\left(\frac{1}{x}\right)+C[/MATH]

 

[pka]

I have the function f-infinity, -1) -> R $\displaystyle f(x)=\frac{1}{x\sqrt{x^{2}-1}}$
I have to find the primitives of this function.So I need to calculate $\displaystyle \int \frac{1}{x\sqrt{x^{2}-1}}$
I solved it with u=sqrt(x^2 - 1) and I got $\displaystyle arctan\sqrt{x^{2}-1} + C$ but the right answer is arcsin(1/x) + C, why?

First, if $\displaystyle x<-1$ then $\displaystyle \sqrt{x^2}=-x$.
\(\displaystyle \begin{align*}\frac{d}{dx}\arcsin\left(x^{-1}\right)&=\frac{\tfrac{d}{dx}\left(x^{-1}\right)}{\sqrt{1-\left(x^{-1}\right)^2}} \\&=\frac{-\left(x^{-2}\right)}{\sqrt{1-\left(x^{-2}\right)}}\\&=\frac{-1}{x^2\sqrt{1-\left(x^{-2}\right)}}\\&=\frac{1}{x\sqrt{x^2-1}} \end{align*}\)

 

[Vali]

It appears to me that you and I assumed $1<x$, and so we want:

[MATH]-\int \frac{1}{x\sqrt{x^2-1}}\,dx=-\arctan(\sqrt{x^2-1})+C[/MATH]
And then we could observe that:

[MATH]\arctan\left(\frac{1}{\sqrt{x^2-1}}\right)=\frac{\pi}{2}-\arctan(\sqrt{x^2-1})[/MATH]
Hence:

[MATH]-\int \frac{1}{x\sqrt{x^2-1}}\,dx=\arctan\left(\frac{1}{\sqrt{x^2-1}}\right)+C[/MATH]
And this is equivalent to:

[MATH]-\int \frac{1}{x\sqrt{x^2-1}}\,dx=\arcsin\left(\frac{1}{x}\right)+C[/MATH]

Why the answer C is not right then ?How to know at exam which answer to pick?
Thanks!

 

[Steven G]

Why the answer C is not right then ?How to know at exam which answer to pick?
Thanks!

From how I see it, MarkFL never got C as an answer. Please look more closely at his work.

 

[MarkFL]

From how I see it, MarkFL never got C as an answer. Please look more closely at his work.

My work wasn't really satisfactory. Let's try again. and state:

[MATH]u=-x\implies du=-dx[/MATH]
Hence

[MATH]F(u)=\int\frac{1}{u\sqrt{u^2-1}}\,du=\arctan(\sqrt{u^2-1})+C=-\arctan\left(\frac{1}{\sqrt{u^2-1}}\right)+C[/MATH]
Thus:

[MATH]F(u)=\arcsin\left(\frac{1}{-u}\right)+C[/MATH]
Hence:

[MATH]F(x)=\arcsin\left(\frac{1}{x}\right)+C[/MATH]