# Thread: help with intergration by recogbition

1. ## help with intergration by recogbition

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

2. Originally Posted by dustinbehemoth
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
What do you get for the derivative?

3. the derivative of g(x) was
(pi)(10-x)sin(pi)x - cos(pi)x

4. Originally Posted by dustinbehemoth
the derivative of g(x) was
(pi)(10-x)sin(pi)x - cos(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.

5. found a heaps easier(may just be better explained) way but thanks anyway