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

Printable View

- Jul 29th 2010, 12:52 AMdustinbehemothhelp 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 - Jul 29th 2010, 01:06 AMmr fantastic
- Jul 29th 2010, 01:09 AMdustinbehemoth
the derivative of g(x) was

(pi)(10-x)sin(pi)x - cos(pi)x - Jul 29th 2010, 05:19 AMmr fantastic
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. - Jul 30th 2010, 12:09 AMdustinbehemoth
found a heaps easier(may just be better explained) way but thanks anyway