oh geeez.... this is a tuff one!! i will still give it a try!!!
∫e^(2x)sin²xdx sin²x = 1/2(1 - cos2x)
∫e^(2x)[1/2(1 - cos2x)]dx let u=2x so du=2dx
1/2∫e^(u)[1/2(1 - cosu)]du
1/4∫e^(u)(1 - cosu)du
1/4∫e^(u)du - 1/4∫e^(u)cosudu
-1/4(∫e^(u)cosudu - ∫e^(u)du) use integration by parts on ∫e^(u)cosudu
f(u) = e^(u) g'(u) = cosu
f'(u) = e^(u) g(u) = sinu
∫f(u)g'(u)du = f(u)g(u) - ∫g(u)f'(u)du
-1/4[(e^(u)sinu - ∫e^(u)sinudu) - ∫e^(u)du] use integration by parts on ∫e^(u)sinudu
f(u) = e^(u) g'(u) = sinu
f'(u) = e^(u) g(u) = -cosu
-1/4[{e^(u)sinu - (-e^(u)cosu + ∫e^(u)cosudu)} - ∫e^(u)du]
-1/4[(e^(u)sinu + e^(u)cosu - ∫e^(u)cosudu) - ∫e^(u)du]
observe that ∫e^(u)cosudu = e^(u)sinu + e^(u)cosu - ∫e^(u)cosudu so
∫e^(u)cosudu + ∫e^(u)cosudu = e^(u)sinu + e^(u)cosu
2∫e^(u)cosudu = e^(u)sinu + e^(u)cosu
∫e^(u)cosudu = 1/2(e^(u)sinu + e^(u)cosu)
finally...
-1/4{[1/2(e^(u)sinu + e^(u)cosu)] - e^(u)}
you can plug u=2x back in there if ya want. i may have done it all wrong or there may be an easier way of doing it who knowns???
]]>I have to solve:
Volume= ∫ π (e^2x * (sinx)^2 * dx ,0 ,π
I know that the integral of e^2x = 1/2 e^2x.... but I don't know what to do with (sinx)^2?
]]>(e^x sinx)^2
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