积分问题
题目如图片
令t=tan(x/2),则dx=(2/(1+t^2))dt,cosx=(1-t^2)/(1+t^2) ∫dx/(1+(cosx)^2) =∫(2/(1+t^2))dt/[1+[(1-t^2)/(1+t^2)]^2] =∫(1+t^2)dt/(1+t^4) =1/2{∫dt/[(t-√2/2)^2+1/2]+∫dt/[(t+√2/2)^2+1/2]} =1/2{√2arctan(√2t-1)+√2arctan(√2t+1)}+c =√2/2[arctan(√2t-1)+arctan(√2t+1)]+c ∫dx/(1+(cosx)^2) =∫(1+t^2)dt/(1+t^4) =√2/2[arctan(√2t-1)+arctan(√2t+1)]| =√2/2[arctan(√2-1)+arctan(√2+1)]-√2/2[arctan(-1)+arctan(1)] =√2/2[arctan(√2-1)+arctan(√2+1)] =√2/2[π/8+3π/8] =√2π/4。
题目错误,为定积分问题,答为一个值
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