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echy5555
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Homework Statement
The integral of sin(x)*cos(x) either equals -1/2*cos^2(x) or 1/2*sin^2(x). Which is it? It can't be both, right?
Homework Equations
[tex]\int[/tex]udv=u*v-[tex]\int[/tex]vdu
The Attempt at a Solution
integration by parts:
u=cos(x) dv=sin(x)dx
du=-sin(x)dx v=-cos(x)
-cos^2(x)-[tex]\int[/tex]sin(x)*cos(x)
so:
2*[tex]\int[/tex]sin(x)*cos(x)=cos^2(x)
but can't it also be done this way?
u=sin(x) dv=cos(x)dx
du=cos(x)dx v=sin(x)
sin^2(x)-[tex]\int[/tex]sin(x)*cos(x)
so:
2*[tex]\int[/tex]sin(x)*cos(x)=sin^2(x)