# Question regarding integration

1. Sep 12, 2011

### olliepower

1. The problem statement, all variables and given/known data
Calculate the integral:

from 0 to pi/2
I=$\int$sin x {2f′(cos x) − 1} dx

When f(0) = 2 and f(1) = 4

2. Relevant equations

3. The attempt at a solution

integral from zero to pi/2 of 2$\int$sin(x){f'(Cos (x) -1}dx

u = cos (x)
du = -sinx dx

(Here is where I run into problems....i do not know know to do with the -1)

1. We insert the limits of integration into the U = cos (x) and get

0 = cos (pi/2)
1 = cos (1)

so now write

-2$\int$f'(u)-1du (limits of integration are from 1 to zero

flip limits of integration
2$\int$ f'u-1du (limits of integration are zero to 1)

integrate

2(F(1)-1) - (2(F(0)-0)
2(4-1) - 2(2-0)
6 - 4 = 2.

I tried 2 in my HM system and it is incorrect. What am I doing wrong?

2. Sep 12, 2011

### dynamicsolo

OK, when you factored out the '2', you forgot to factor it out of the second term of 1 ....