Question

不定积分问题 sin^2(3x)cosxdx dx/(a^2-x^2)^3/2 dx/x^2*根号（x^2-a^2）要详

Answer 1

不定积分问题 ①∫sin²(3x)cosxdx；② ∫dx/(a²-x²)^(3/2) ；③∫dx/x²√(x²-a²）要详细过程

①∫sin²(3x)cosxdx=∫(3sinx-4sin³x)²cosxdx=∫[9sin²x-24sin⁴x+16(sinx)^6]d(sinx)

=3sin³x-(24/5)(sinx)^5+(16/7)(sinx)^7+C

[此处用了三角公式：sin3x=3sinx-4sin³x]

② ∫dx/(a²-x²)^(3/2)=∫dx/[(a²-x²)√(a²-x²)]=∫dx/{(a²-x²)a√[1-(x/a)²]}

令x/a=sinu,则x=asinu,dx=acosudu,代入上式得：

=∫acosudu/[a²-a²sin²u)a√(1-sin²u)=∫du/(a²cos²u)=(1/a²)tanu+C=(1/a²)[x/√(a²-x²)]+C

③∫dx/x²√(x²-a²)=(1/a)∫dx/{x²√[(x/a)²-1]}

令x/a=secu,则x=asecu,dx=asecutanudu,代入上式得：

=∫secutanudu/[a²sec²u√(sec²u-1)]=∫secutanudu/a²sec²utanu=(1/a²)∫du/secu=(1/a²)∫cosudu

=(1/a²)sinu+C=(1/a²)[√(x²-a²)]/x+C=[√(x²-a²)]/a²x+C