T tessec New member Joined Sep 21, 2006 Messages 4 Sep 27, 2006 #1 problem: integral of (x^5)cos(x^3) work: f(x)=cos(x^3) g'(x)=x^5 f'(x)=-3(x^2)sin(x^3) g(x)=(1/6)(x^6) => (1/6)(x^6)(cos(x^3))+(1/2)int[(x^8)(sin(x^3)]
problem: integral of (x^5)cos(x^3) work: f(x)=cos(x^3) g'(x)=x^5 f'(x)=-3(x^2)sin(x^3) g(x)=(1/6)(x^6) => (1/6)(x^6)(cos(x^3))+(1/2)int[(x^8)(sin(x^3)]
skeeter Elite Member Joined Dec 15, 2005 Messages 3,204 Sep 27, 2006 #2 problem: integral of (x^5)cos(x^3) Click to expand... \(\displaystyle \L u = x^3 ...... dv = x^2 cos(x^3) dx\) \(\displaystyle \L du = 3x^2 dx ...... v = \frac{1}{3}sin(x^3)\) \(\displaystyle \L uv - \int v du\) \(\displaystyle \L \frac{x^3}{3}sin(x^3) - \int x^2 sin(x^3) dx =\) \(\displaystyle \L \frac{x^3}{3}sin(x^3) + \frac{1}{3}cos(x^3) + C\)
problem: integral of (x^5)cos(x^3) Click to expand... \(\displaystyle \L u = x^3 ...... dv = x^2 cos(x^3) dx\) \(\displaystyle \L du = 3x^2 dx ...... v = \frac{1}{3}sin(x^3)\) \(\displaystyle \L uv - \int v du\) \(\displaystyle \L \frac{x^3}{3}sin(x^3) - \int x^2 sin(x^3) dx =\) \(\displaystyle \L \frac{x^3}{3}sin(x^3) + \frac{1}{3}cos(x^3) + C\)