Question

Why does the cos x get removed when integrating this? Help!?

Answer 1

Because of the chain rule

Explanation:

Recall that the chain rule states that for `f (g (x)), d/(dx)f (g l x) = g'(x)f'(g (x))` . Further, recall that the derivative of `e^x` is simply `e^x`. Thus, for `e^u`, where `u=sin (x)`, the derivative is `(du)/(dx)*e^u` since cos x is the derivative of sin x...

`d/(dx) (e^sin (x)+c = cos (x)e^sin (x)`

Do note that in your problem, you must still evaluate the integral at the endpoints mentioned to finish calculating the area.

Answer 2

The area is `e - 1` square units.

Explanation:

Yes, start by finding the anti-derivative. Our bounds of integration will be `[0, pi/2]`.

`I = int_0^(pi/2) cosxe^(sinx)`

Now let `u = sinx`. Then `du = cosxdx`. We change the bounds of integration accordingly.

`I = int_0^1 cosxe^u * (du)/(cosx)`

`I = int_0^1 e^u du`

`I = e^1 - e^0 = e - 1`

The area under the curve is `e - 1`.

Hopefully this helps!