Question

求大神帮忙看看高数题

希望能有详细步骤，感觉混淆了

Answer 1

当t属于（π/2，π）时，与前面那个区间只差一个负号

Answer 2

令x＝a×sect试试

Answer 3

三角函数代换法！
设x=asecu，则dx=asecutanudu，x²-a²=a²(sec²u-1)=a²tan²u，
√(x²-a²)=atanu，secu=x/a, tanu=[√(x²-a²)]/a.
代入原式得:
∫[√(x²-a²)]dx
=a²∫tan²usecudu
=a²∫secu(sec²u-1)du
=a²[∫sec³udu-∫secudu]
=a²[(1/2)secutanu+(1/2)ln(secu+tanu)-ln(secu+tanu)]+lnc₁
=a²[(1/2)secutanu-(1/2)ln(secu+tanu)]+lnc₁
=(1/2)[(x√(x²-a²)]-(a²/2)ln[(x/a)+(1/a)√(x²-a²)]+lnc₁
=(1/2){x√(x²-a²)-a²[ln(x+√(x²-a²)-lna]}+lnc₁
=(1/2){x√(x²-a²)-a²ln[x+√(x²-a²)]}+a²ln(ac₁)
=(1/2){x√(x²-a²)-a²ln[x+√(x²-a²)]}+C

追答

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