Bonjour,
a. u'v = (uv)'-uv' rappel : (uv)' = u'v+uv'
u'v = u'v+uv'-uv'
u'v = u'v
b. [tex]\int\limits^b_a {uv'(t)} \, dt = \int\limits^b_a {[u'v(t)+uv'(t)]} \, dt \ soit [uv(t)]^{b} _{a} =\int\limits^b_a {u'v(t)} \, dt +\int\limits^b_a {uv'(t)} \, dt\\[/tex]
D'après l'égalité précédente : [tex]\int\limits^b_a {u'v(t)} \, dt = [uv(t)]^{b} _{a} -\int\limits^b_a {uv'(t)} \, dt[/tex]
[tex]\int\limits^\frac{\pi }{2} _ {0}tsin(t) \, dt \\\\u = t\\u'=t'\\\\v'=sin(t)dt\\v=-cos(t)\\\\\\\int\limits{t*sin(t)} \, dt\\ t*(-cos(t))-\int\limits {-cos(t)} \, dt\\ -t*cos(t)+\int\limits {cos(t)} \, dt\\-t*cos(t)+sin(t)\\(-t*cos(t)+sin(t)) |^{\frac{\pi }{2} } _{0} \\-\frac{\pi }{2}*cos(\frac{\pi }{2})+sin(\frac{\pi }{2})-(-0*cos(0)+sin(0))\\ = 1[/tex]
3.
[tex]\int\limits^x_1 {ln(t)} \, dt\\ \\u=ln(t)\\v'=1\\u'= \frac{1}{t}\\ v=t\\\\\int\limits {ln(t)} \, dt =ln(t)t-(\int\limits {1} \, dt) \\ ln(t)t-t\\\\(ln(t)*t-t) |^{x}_{1}\\ ln(x)*x-x+1[/tex]
La fonction F représente la primitive
