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1. 设 x = 2sint,则 dx = 2cost*dt。那么,原来的定积分就变换成:
=∫(2sint)^2 * (2cost)*(2cost) * dt,积分下限:t = 0,积分上限:t = π/2
=∫(2sint * 2cost)^2 * dt
=∫(2sint*cost)^2 * 4 * dt
=∫[sin(2t)]^2 * 4 * dt
=∫2*[sin(2t)]^2 * (2 * dt)
=∫[1 - cos(4t)] * (2 * dt)
=∫2*dt - ∫cos(4t)* (1/2) * (4 * dt)
=2t - 1/2 * ∫cos(4t)*d(4t)
=2t - 1/2 * sin(4t)
=2π - 1/2 * [sin(2π) - sin0]
=2π
2. =∫x^2 * e^(x^2) * (x * dx),设 t = x^2,则积分下限:t = 0,积分上限:t = √2
=1/2 * ∫t * e^t * dt
=1/2 * [t * e^t - ∫e^t * dt]
=1/2 * [t * e^t - e^t]
=1/2 * (t - 1)*e^t
=1/2 * [(√2 - 1)*e^(√2) - (0 - 1)*e^0]
=1/2 * [(√2 - 1)*e^(√2) + 1]
=∫(2sint)^2 * (2cost)*(2cost) * dt,积分下限:t = 0,积分上限:t = π/2
=∫(2sint * 2cost)^2 * dt
=∫(2sint*cost)^2 * 4 * dt
=∫[sin(2t)]^2 * 4 * dt
=∫2*[sin(2t)]^2 * (2 * dt)
=∫[1 - cos(4t)] * (2 * dt)
=∫2*dt - ∫cos(4t)* (1/2) * (4 * dt)
=2t - 1/2 * ∫cos(4t)*d(4t)
=2t - 1/2 * sin(4t)
=2π - 1/2 * [sin(2π) - sin0]
=2π
2. =∫x^2 * e^(x^2) * (x * dx),设 t = x^2,则积分下限:t = 0,积分上限:t = √2
=1/2 * ∫t * e^t * dt
=1/2 * [t * e^t - ∫e^t * dt]
=1/2 * [t * e^t - e^t]
=1/2 * (t - 1)*e^t
=1/2 * [(√2 - 1)*e^(√2) - (0 - 1)*e^0]
=1/2 * [(√2 - 1)*e^(√2) + 1]
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1、解:设x=2sint
得:∫x²√(4-x²)dx=∫ 4sin²t*2cost*2cost dt=16∫ sin²t*cos²t dt=16∫sin²2t/4dt=4∫(1-cos4t)/2dt
=2(t-1/4*sin4t)
原式=2[(π/2-1/4*sin2π)-(0-1/4*sin4*0)]=π-1/2
2、解:∫x³e^x²dx=x²/2*e^x²-∫x*e^x²dx=x²/2*e^x²-1/2*e^x²
原始=3/2*e^4-1/2
得:∫x²√(4-x²)dx=∫ 4sin²t*2cost*2cost dt=16∫ sin²t*cos²t dt=16∫sin²2t/4dt=4∫(1-cos4t)/2dt
=2(t-1/4*sin4t)
原式=2[(π/2-1/4*sin2π)-(0-1/4*sin4*0)]=π-1/2
2、解:∫x³e^x²dx=x²/2*e^x²-∫x*e^x²dx=x²/2*e^x²-1/2*e^x²
原始=3/2*e^4-1/2
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