求∫sin^2(x)[sin^4(x)+ln(3+x)/(3-x)]dx在[-π/2,π/2]上的定积分^表示次幂
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求∫sin^2(x)[sin^4(x)+ln(3+x)/(3-x)]dx在[-π/2,π/2]上的定积分^表示次幂
求∫sin^2(x)[sin^4(x)+ln(3+x)/(3-x)]dx在[-π/2,π/2]上的定积分
^表示次幂
求∫sin^2(x)[sin^4(x)+ln(3+x)/(3-x)]dx在[-π/2,π/2]上的定积分^表示次幂
ln(3+x)/(3-x) 是奇函数,∫[-π/2,π/2] sin^2(x) * ln(3+x)/(3-x) dx = 0
∫[-π/2,π/2] (sinx)^6 dx
= 2 ∫[0,π/2] (sinx)^6 dx 有公式
= 2 * 5*3/(6*4*2) * (π/2)
= 5π/16
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