# Integral by substitution

• November 18th 2010, 02:52 PM
juanma101285
Integral by substitution
Hi, I have the following problem but I do not know what to do...what substitution should I choose? Thanks in advance!

Using a substitution of the form $y=a+b sinh(t)$, or otherwise, find polynomial $p(y)$ such that $\int \sqrt{10+y^2-2y} dy = +p(y) \sqrt{10+y^2-2y} + C$.
• November 18th 2010, 03:00 PM
TheCoffeeMachine
What does the plus that precedes p(y) indicate? If you want the integral:

Write $y^2-2y+10 = (y-1)^2+9$ and let $y+1 = 3\tan{t}$.
You will need to find the integral $\sec^3{t}$, for which you should see here.
• November 18th 2010, 04:12 PM
juanma101285
Hi,

I am not sure...I guess it means that it is positive? It does not say anything else.

I have done it but I have got the wrong result, I think.

I got
$9/2 arsinh((y-1)/3)+(1/2)((y-1)/3) \sqrt{10+y^2+2y} + C$

I used a different substitution though, $y=1+3 sinh(t)$, with $dy=3 cosh(t) dt$.

A summary of my calculations is here:

$\int \sqrt{10+y^2-2y} dy = \int \sqrt{(y-1)^2+3^2} dy$
$= \int \sqrt{(3 sinh(t))^2+3^2} 3 cosh(t) dt$
$= 9 \int \sqrt{1+sinh^2(t)} cosh(t) dt = 9 \int cosh^2(t) dt$
$= 9/2 \int (1 + cosh(2t)) dt = 9/2 (t + sinh(2t)/2) + C = 9/2 (t sinh(t) + cosh(t)) + C$

Substituting back to y:

$= 9/2 * arsinh((y-1)/3) + 1/2 * ((y-1)/3) \sqrt{10+y^2-2y} + C$

(using the fact that $cosh(t) = \sqrt{1+sinh^2(t)} = \sqrt{10+y^2-2y}$
• November 18th 2010, 04:24 PM
Prove It
Quote:

Originally Posted by juanma101285
Hi, I have the following problem but I do not know what to do...what substitution should I choose? Thanks in advance!

Using a substitution of the form $y=a+b sinh(t)$, or otherwise, find polynomial $p(y)$ such that $\int \sqrt{10+y^2-2y} dy = +p(y) \sqrt{10+y^2-2y} + C$.

$\displaystyle \sqrt{y^2 - 2y + 10} = \sqrt{y^2 - 2y + (-1)^2 - (-1)^2 + 10}$

$\displaystyle = \sqrt{(y-1)^2 + 9}$.

Now make the substitution $\displaystyle y - 1 = 3\sinh{t}$ so that $\displaystyle dy = 3\cosh{t}\,dt$.

Then your integral becomes

$\displaystyle \int{\sqrt{(y-1)^2 + 9}\,dy} = \int{\sqrt{(3\sinh{t})^2 + 9}\,3\cosh{t}\,dt}$

$\displaystyle = \int{3\cosh{t}\sqrt{9\sinh^2{t} + 9}\,dt}$

$\displaystyle = \int{3\cosh{t}\sqrt{9(\sinh^2{t} + 1)}\,dt}$

$\displaystyle = \int{3\cosh{t}\sqrt{9\cosh^2{t}}\,dt}$

$\displaystyle = \int{3\cosh{t}\cdot 3\cosh{t}\,dt}$

$\displaystyle = \int{9\cosh^2{t}\,dt}$

$\displaystyle = \int{9\left(\frac{1}{2}\cosh{2t} + \frac{1}{2}\right)\,dt}$

$\displaystyle = \int{\frac{9}{2}\cosh{2t} + \frac{9}{2}\,dt}$

$\displaystyle = \frac{9}{4}\sinh{2t} + \frac{9}{2}t + C$

$\displaystyle = \frac{9}{2}\sinh{t}\cosh{t} + \frac{9}{2}t + C$

$\displaystyle = \frac{9}{2}\sinh{t}\sqrt{1 + \sinh^2{t}} + \frac{9}{2}t + C$

$\displaystyle = \frac{9}{2}\left(\frac{y-1}{3}\right)\sqrt{1 + \left(\frac{y-1}{3}\right)^2} + \frac{9}{2}\sinh^{-1}\left(\frac{y-1}{3}\right) + C$

$\displaystyle = \frac{3(y-1)}{2}\frac{\sqrt{y^2 - 2y + 10}}{\sqrt{9}} + \frac{9}{2}\sinh^{-1}\left(\frac{y-1}{3}\right) + C$

$\displaystyle = \frac{y-1}{2}\sqrt{y^2 - 2y + 10} + \frac{9}{2}\sinh^{-1}\left(\frac{y-1}{3}\right) + C$.
• November 18th 2010, 04:42 PM
juanma101285
Thanks! :)