Inverse Trigonometric Function Integral?

**Torcida1911** · March 3rd 2010, 10:06 PM

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.

**Prove It** · March 4th 2010, 12:15 AM

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 $x - 3 = \cosh{t}$ .

**Torcida1911** · March 4th 2010, 06:33 AM

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) ?

thank you very much for your answer!!

**TheEmptySet** · March 4th 2010, 07:42 AM

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) ?

thank you very much for your answer!!

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

This gives

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

**Miss** · March 4th 2010, 08:26 AM

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.

**Prove It** · March 4th 2010, 04:00 PM

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) ?

thank you very much for your answer!!

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".

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