Answer:
[tex]\int\limits^\pi_0 {\sin^2(x)\cos(x)} \, dx=0[/tex]
Step-by-step explanation:
So we have the integral:
[tex]\int\limits^\pi_0 {\sin^2(x)\cos(x)} \, dx[/tex]
To evaluate this integral, we can use u-substitution. Remember that the derivative of sin(x) is cos(x). So, let u equal sin(x):
[tex]u=\sin(x)[/tex]
Take the derivative of u:
[tex]\frac{du}{dx}=\cos(x)[/tex]
Multiply both sides by dx:
[tex]du=\cos(x)dx[/tex]
So, we can substitute cos(x) x for du.
We can also substitute sin(x) for u. Thus:
So, our integral is now:
[tex]\int\limits^\pi_0 {\sin^2(x)(\cos(x)} \, dx)\\[/tex]
This is equal to:
[tex]=\int\limits^\pi_0 {u^2} \, du[/tex]
However, we also must change our bounds of integration. To do so, substitute in the lower and upper bound into u. So:
[tex]u=\sin(x)\\u=\sin(0)=0[/tex]
And:
[tex]u=\sin(x)\\u=\sin(\pi)=0[/tex]
Therefore, our integral with our new bounds is:
[tex]=\int\limits^0_0 {u^2} \, du[/tex]
Now, note that the integral has the same upper bound and lower bound. Therefore, this means that our integral is going to be 0 since with the same bounds, there will be no area.
Therefore, our answer is 0:
[tex]\int\limits^0_0 {u^2} \, du=0[/tex]
And we're done!
