# Thread: Inverse Trigonometric Function Integral?

1. ## Inverse Trigonometric Function Integral?

If you could help me figure this integral out I would be very grateful.

top limit: 6
bottom limit: 4

of dx / (x-3)sqrt(x^2-6x+8)

The question involves the arc graphs and whatnot. I'm just really confused.
Sorry if this is a little messy, I'm not experienced with latex. any help would be appreciated, thanks.

Im not even sure where to begin. I know i must make a u substitution but I dont know what with.

2. Originally Posted by Torcida1911
If you could help me figure this integral out I would be very grateful.

top limit: 6
bottom limit: 4

of dx / (x-3)sqrt(x^2-6x+8)

The question involves the arc graphs and whatnot. I'm just really confused.
Sorry if this is a little messy, I'm not experienced with latex. any help would be appreciated, thanks.

Im not even sure where to begin. I know i must make a u substitution but I dont know what with.
So this is $\int_4^6{\frac{1}{(x - 3)\sqrt{x^2 - 6x + 8}}\,dx}$

$= \int_4^6{\frac{1}{(x - 3)\sqrt{x^2 - 6x + (-3)^2 - (-3)^2 + 8}}\,dx}$

$= \int_4^6{\frac{1}{(x - 3)\sqrt{(x - 3)^2 - 1}}\,dx}$.

Now try making the substitution $x - 3 = \cosh{t}$.

3. Originally Posted by Prove It
So this is $\int_4^6{\frac{1}{(x - 3)\sqrt{x^2 - 6x + 8}}\,dx}$

$= \int_4^6{\frac{1}{(x - 3)\sqrt{x^2 - 6x + (-3)^2 - (-3)^2 + 8}}\,dx}$

$= \int_4^6{\frac{1}{(x - 3)\sqrt{(x - 3)^2 - 1}}\,dx}$.

Now try making the substitution $x - 3 = \cosh{t}$.
Alright, I'll try this.

$= \int_4^6{\frac{1}{(cosht)\sqrt{(cosht)^2 - 1}}\,dx}$.
= $= \int_4^6{\frac{1}{(cosht)\sqrt{(cosht)^2 - (cosht^2 + sinht^2)}}\,dx}$.
= $= \int_4^6{\frac{1}{(cosht)\sqrt{(-sinht^2)}}\,dx}$.

Ah..I've made an error here. I can't square root that..

Maybe I should set x-3 = u? that way I'll get a similar formula to that of
d/dx (arcsecx) = 1 / |x|sqrt(x^2 -1) ?

4. Originally Posted by Torcida1911
Alright, I'll try this.

$= \int_4^6{\frac{1}{(cosht)\sqrt{(cosht)^2 - 1}}\,dx}$.
= $= \int_4^6{\frac{1}{(cosht)\sqrt{(cosht)^2 - (cosht^2 + sinht^2)}}\,dx}$.
= $= \int_4^6{\frac{1}{(cosht)\sqrt{(-sinht^2)}}\,dx}$.

Ah..I've made an error here. I can't square root that..

Maybe I should set x-3 = u? that way I'll get a similar formula to that of
d/dx (arcsecx) = 1 / |x|sqrt(x^2 -1) ?

Try setting $x-3=\sec(t) \implies dx=\sec(t)\tan(t)dt$

This gives

$\int_4^6{\frac{1}{(x - 3)\sqrt{(x - 3)^2 - 1}}\,dx}=\int \frac{\sec(t)\tan(t)}{\sec(t)\sqrt{\sec^2(t)-1}}dt=\int dt$

5. Originally Posted by Torcida1911
If you could help me figure this integral out I would be very grateful.

top limit: 6
bottom limit: 4

of dx / (x-3)sqrt(x^2-6x+8)

The question involves the arc graphs and whatnot. I'm just really confused.
Sorry if this is a little messy, I'm not experienced with latex. any help would be appreciated, thanks.

Im not even sure where to begin. I know i must make a u substitution but I dont know what with.
They gave you the solution.
But the problem is they did not notice that the integral is improper.

6. Originally Posted by Torcida1911
Alright, I'll try this.

$= \int_4^6{\frac{1}{(cosht)\sqrt{(cosht)^2 - 1}}\,dx}$.
= $= \int_4^6{\frac{1}{(cosht)\sqrt{(cosht)^2 - (cosht^2 + sinht^2)}}\,dx}$.
= $= \int_4^6{\frac{1}{(cosht)\sqrt{(-sinht^2)}}\,dx}$.

Ah..I've made an error here. I can't square root that..

Maybe I should set x-3 = u? that way I'll get a similar formula to that of
d/dx (arcsecx) = 1 / |x|sqrt(x^2 -1) ?

You have forgotten that if $x - 3 = \cosh{t}$ then $dx = \sinh{t}\,dt$.

So the integral actually becomes:

$\int{\frac{1}{\cosh{t}\sqrt{\cosh^2{t} - 1}}\,\sinh{t}\,dt}$

Now since $\cosh^2{t} - \sinh^2{t} = 1$ that means $\cosh^2{t} - 1 = \sinh^2{t}$.

$= \int{\frac{\sinh{t}}{\cosh{t}\sqrt{\sinh^2{t}}}\,d t}$

$= \int{\frac{\sinh{t}}{\cosh{t}\sinh{t}}\,dt}$

$= \int{\frac{1}{\cosh{t}}\,dt}$

$= \int{\textrm{sech}\,{t}\,dt}$.

I suggest you consult your tables of integrals to evaluate this integral.

For an alternative method,

integral[1/((x - 3)*Sqrt[(x - 3)^2 - 1])] - Wolfram|Alpha

Click "Show Steps".