求详细的积分步骤
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求定积分【0，π/4】∫[(2acosθ)⁴/4]dθ
解：原式=【0，π/4】4a⁴∫cos⁴θdθ=【0，π/4】4a⁴∫[(1+cos2θ)/2]²dθ
=【0，π/4】a⁴∫[(1+2cos(2θ)+cos²2θ)dθ=【0，π/4】a⁴[∫dθ+∫cos2θd(2θ)+∫cos²2θdθ
=【0，π/4】a⁴{∫dθ+∫cos(2θ)d(2θ)+(1/8)∫[(1+cos4θ)]d(4θ)}
=【0，π/4】a⁴[θ+sin(2θ)+(1/8)(4θ+sin4θ)]
=【0，π/4】a⁴[3θ/2+sin(2θ)+(1/8)sin(4θ)]=(3π/8+1)a⁴
解：原式=【0，π/4】4a⁴∫cos⁴θdθ=【0，π/4】4a⁴∫[(1+cos2θ)/2]²dθ
=【0，π/4】a⁴∫[(1+2cos(2θ)+cos²2θ)dθ=【0，π/4】a⁴[∫dθ+∫cos2θd(2θ)+∫cos²2θdθ
=【0，π/4】a⁴{∫dθ+∫cos(2θ)d(2θ)+(1/8)∫[(1+cos4θ)]d(4θ)}
=【0，π/4】a⁴[θ+sin(2θ)+(1/8)(4θ+sin4θ)]
=【0，π/4】a⁴[3θ/2+sin(2θ)+(1/8)sin(4θ)]=(3π/8+1)a⁴
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