How do I evaluate this integral, given a table of values for f(x) and f'(x)? The problem gives me a list of values for $f(x)$ and $f'(x)$. I tried solving the problem using integration by parts, allowing $u=\sin(2x), u'=2\cos(2x), v'=f'(\cos(2x))$, and $v=f(\cos(2x))$. I use parts again and come up with $v'=f(\cos(2x))$ and I think this is where I'm stuck because I'm not sure how to integrate that.. Can I just turn it into $v=\cos(2x)$?? Am I even taking the correct approach using parts?
I know that when solving an integration problem just using functions of $x$ and values, that you can construct a riemann sum and solve it that way, but the incorporation of $f(x)$ and $f'(x)$ is throwing me off. 
Here is a picture of the problem:

 A: I totally don't know if this is right what I'm doing, but I'd like to know if my approach is correct.
If we straightly substitute in 
$$u = \cos(2x)$$
We'd have
$$du = -2\sin(2x)\; dx $$
Or equivalently
$$-\frac{1}{2}\; \text{d}u = \sin(2x)\; \text{d}x $$
When $x=\pi / 2$, then $u = \cos(2\pi/2) = \cos(\pi) = -1$ and when $x = 0$, $u = \cos(0) =1$. 
So we can transform the whole integral from
$$\int_{0}^{\pi/2} \sin(2x) \cdot f'(\cos(2x)) \; \text{d}x$$
to
$$\int_{1}^{-1} -\frac{1}{2}\cdot f'(u) \; \text{d}u$$
And I guess it will help that we know $f(1)$, $f(-1)$, $f'(-1)$ and $f'(1)$.
Since then we have:
$$\int_{1}^{-1} -\frac{1}{2}\cdot f'(u) \; \text{d}u = -\frac{1}{2} \int_1^{-1} f'(u) \; \text{d}u = -\frac{1}{2}\left(f(-1) - f(1) \right) $$
And we can compute that constant now because we know the function values.
However it seems awfully suspicious that this answer doesn't need any function values of $f'()$ at all.. Hm.. 
EDIT:
Weird, even the formula on wikipedia (https://en.wikipedia.org/wiki/Integration_by_substitution#Definition) straightly says
$$\int_{\varphi(a)}^{\varphi(b)} f(x) \; \text{d}x = \int_a^b f(\varphi(t))\varphi'(t) \; \text{d}t$$
Working from the right side we get from the original integral
$$\int_{0}^{\pi/2} \sin(2x) \cdot f'(\cos(2x)) \; \text{d}x = -\frac{1}{2}\int_{0}^{\pi/2} -2\sin(2x) \cdot f'(\cos(2x)) \; \text{d}x$$
We can see that in this case 
$$\varphi(x) = \cos(2x)$$
$$\varphi'(x) = -2\sin(2x)$$
$$a = 0 \quad b=\pi/2$$
And therefore the bounds on the left integral must be
$$\varphi(a) = \cos(0) = 1$$
$$\varphi(b) = \cos(2\cdot \pi/2) = -1$$
And now we get the left side as
$$\int_1^{-1} -\frac{1}{2}f'(u) \; \text{d}u = -\frac{1}{2}\left(f(-1) - f(1)\right)  = -\frac{1}{2}(10-7) = -\frac{3}{2}$$
Which is the same as before -.-.
If the upper bound was however $0$, we'd have 
$$\int_1^{0} -\frac{1}{2}f'(u) \; \text{d}u = -\frac{1}{2}\left(f(0) - f(1)\right)  = -\frac{1}{2}(8-7) = -\frac{1}{2}$$ 
Nope, no idea what's wrong. Now I'd also like to learn what the correct answer is.
A: Take $f(cos(2x))=t$.
Then we have $f'(cos(2x))*-sin(2x)*2 dx= dt$
Substitute this into the integral to get $\int -1/2 dt$
The limits will be $f(1)=7$ to $f(-1)=10$, which gives you
$\int_{7}^{10} -1/2 dt = -3/2$
So the required answer is $3/2$.
A: Hermite interpolation
finds the polynomial that
fits a function
given function and
derivative values.
https://en.wikipedia.org/wiki/Hermite_interpolation
Construct the Hermite interpolation
polynomial for your data
and then integrate it
from min to max value.
