神马作文网（cosx^2 ）（e^x）在[0,π]求定积分
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（cosx^2 ）（e^x）在[0,π]求定积分
![（cosx^2 ）（e^x）在[0,π]求定积分](/uploads/image/z/1639764-36-4.jpg?t=%EF%BC%88cosx%5E2+%EF%BC%89%EF%BC%88e%5Ex%EF%BC%89%E5%9C%A8%5B0%2C%CF%80%5D%E6%B1%82%E5%AE%9A%E7%A7%AF%E5%88%86)
∫e^x*cos²x dx when x=0 to π
= (1/2)∫e^x(1+cos2x) dx,simplify the expression first.
= (1/2)∫(e^x+e^x*cos2x) dx
= (1/2)∫e^x dx + (1/2)∫e^x*cos2x dx
For (1/2)∫e^x dx
= 1/2*e^x(0 to π)
= (e^π-1)/2
For (1/2)∫e^x*cos2x dx
Applying the formula ∫e^(ax)*cos(bx) = e^(ax)*[acos(bx)+bsin(bx)]/(a²+b²)
= (1/2)*e^x*(cos2x+2sin2x)/(1+2²)(0 to π)
= (1/10)*[e^π*(cos2π+2sin2π)-e^0*(cos0+2sin0)]
= (e^π-1)/10
So the integral is equal to：
(e^π-1)/2 + (e^π-1)/10
= (3/5)(e^π-1)
= (1/2)∫e^x(1+cos2x) dx,simplify the expression first.
= (1/2)∫(e^x+e^x*cos2x) dx
= (1/2)∫e^x dx + (1/2)∫e^x*cos2x dx
For (1/2)∫e^x dx
= 1/2*e^x(0 to π)
= (e^π-1)/2
For (1/2)∫e^x*cos2x dx
Applying the formula ∫e^(ax)*cos(bx) = e^(ax)*[acos(bx)+bsin(bx)]/(a²+b²)
= (1/2)*e^x*(cos2x+2sin2x)/(1+2²)(0 to π)
= (1/10)*[e^π*(cos2π+2sin2π)-e^0*(cos0+2sin0)]
= (e^π-1)/10
So the integral is equal to：
(e^π-1)/2 + (e^π-1)/10
= (3/5)(e^π-1)
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