# Change of variable in an integral

1. Nov 28, 2017

### Bestfrog

1. The problem statement, all variables and given/known data
A massless string of length 2l connects two hockey pucks that lie on frictionless ice. Aconstant horizontal force F is applied to the midpoint of the string, perpendicular to it (see right figure). How much kinetic energy is lost when the pucks collide, assuming they stick together?

3. The attempt at a solution
Putting a y-axis along the string when it is straight and a x-axis along the direction of the force, the force acting on the puck "at the bottom" along the t-axis is $\frac{F}{2}\cdot tan \theta$.
So, by calculating the work for this puck I obtain $W=\int_{-l}^{0} \frac{F\cdot tan \theta}{2}dy=\int_{-l}{0} \frac{F\cdot tan \theta}{2} d(l\cdot sin\theta)= \int_{\frac{-\pi}{2}}^{0} \frac{Fl}{2}\cdot sin\theta d\theta$.
Can you explaine me how the last passage works?

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2. Nov 28, 2017

### phyzguy

It's a change of variables $y = l \sin(\theta)$.

Since l is constant: $d(l \sin(\theta)) = l d(\sin(\theta)) = l \cos(\theta) d \theta$.

So: $\frac{F \tan(\theta)}{2}d(l \sin(\theta)) = \frac{F l \tan(\theta) \cos(\theta)}{2}d \theta = \frac{F l \sin(\theta)}{2}d \theta$.

Then since $y = l \sin(\theta)$, you need to change the integration limit from $y = -l$ to $\theta = \frac{-\pi}{2}$.