G(x)=(10-x)cos((pi)x)
find an expression for the derivative G'(x)
i got that but then it says
hence show that an antiderivative of f(x) is
F(x)= ((sin((pi)x)) \ (pi)) - G(x)
and f(x) =pi(10-x)sn(pi)x
Let $\displaystyle y =(10-x) \cos ( \pi x)$ .... (1)
Then $\displaystyle \frac{dy}{dx} = \pi (10 - x) \sin (\pi x) - \cos (\pi x)$ .... (2)
Integrate both sides of (2) with respect to x:
$\displaystyle y = \int \pi (10 - x) \sin (\pi x) - \cos (\pi x) \, dx = \int \pi (10 - x) \sin (\pi x) \, dx - \int \cos (\pi x) \, dx $.
Substitute from (1):
$\displaystyle (10-x) \cos ( \pi x) = \int \pi (10 - x) \sin (\pi x) \, dx - \int \cos (\pi x) \, dx $.
Make the required integral the subject etc.