Ask Me Help Desk - Integration for Trough

Are you familiar with integration at all? Since this is a fluid force problem, I assumed

perhaps you were in a calculus II class. That is where this stuff is commonly covered.

I assumed wrong?

Anyway, bring the 2 out and rewrite as:

$124.8\int_{0}^{1}x\sqrt{1-x^{2}}dx$

Now, do you know trig sub? I reckon not?

Use the substitution $x=sin(t), \;\ dx=cos(t)dt$

By doing this, it changes the limits of integration from x=0.. 1 to 0.. Pi/2

$124.8\int_{0}^{\frac{\pi}{2}}sin(t)\sqrt{1-sin^{2}(t)}cos(t)dt$

As you hopefully know, $cos^{2}(t)=1-sin^{2}(t)$

Then, we get $124.8\int_{0}^{\frac{\pi}{2}}sin(t)cos^{2}(t)dt$

We can use the identity $sin(t)cos^{2}(t)=\frac{sin(3t)}{4}+\frac{sin(t)}{4 }$

Now, these are straight forward and we can integrate each separately and finish. Can you do that?