Question

How do you do this question?

Answer 1

Answer:

0

Step-by-step explanation:

∫ sin²(x) cos(x) dx

If u = sin(x), then du = cos(x) dx.

∫ u² du

⅓ u³ + C

⅓ sin³(x) + C

Evaluate between x=0 and x=π.

⅓ sin³(π) − ⅓ sin³(0)

0

Answer 2

Answer:

$\int\limits^\pi_0 {\sin^2(x)\cos(x)} \, dx=0$

Step-by-step explanation:

So we have the integral:

$\int\limits^\pi_0 {\sin^2(x)\cos(x)} \, dx$

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):

$u=\sin(x)$

Take the derivative of u:

$\frac{du}{dx}=\cos(x)$

Multiply both sides by dx:

$du=\cos(x)dx$

So, we can substitute cos(x) x for du.

We can also substitute sin(x) for u. Thus:

So, our integral is now:

$\int\limits^\pi_0 {\sin^2(x)(\cos(x)} \, dx)$

This is equal to:

$=\int\limits^\pi_0 {u^2} \, du$

However, we also must change our bounds of integration. To do so, substitute in the lower and upper bound into u. So:

$u=\sin(x)\\u=\sin(0)=0$

And:

$u=\sin(x)\\u=\sin(\pi)=0$

Therefore, our integral with our new bounds is:

$=\int\limits^0_0 {u^2} \, du$

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:

$\int\limits^0_0 {u^2} \, du=0$

And we're done!