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定积分第四题 求大神
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u=π-x
∫(π/2,π)(sinx)^ndx=∫(π/2,0)(sinu)^nd(-u)=∫(0,π/2)(sinu)^ndu=∫(0,π/2)(sinx)^ndx
∫(0,π/2)(sinx)^ndx+∫(π/2,π)(sinx)^ndx=∫(0,π)(sinx)^ndx
∫(0,π)(sinx)^ndx=2∫(0,π/2)(sinx)^ndx
∫(π/2,π)(sinx)^ndx=∫(π/2,0)(sinu)^nd(-u)=∫(0,π/2)(sinu)^ndu=∫(0,π/2)(sinx)^ndx
∫(0,π/2)(sinx)^ndx+∫(π/2,π)(sinx)^ndx=∫(0,π)(sinx)^ndx
∫(0,π)(sinx)^ndx=2∫(0,π/2)(sinx)^ndx
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