这两道题怎么解，有过程解析给我一下，谢谢

Question

如题所述

Answer 1

见图

Answer 2

∫ x√(1-x²) dx
=1/2*∫ √(1-x²) dx²
=-1/2*∫ (1-x²)^(1/2) d(1-x²)
=-1/2*2/3*(1-x²)^(3/2)+C
=-1/3*(1-x²)^(3/2)+C
∫dx/cos²x√tanx

令t=√tanx，那么tanx=t²，x=arctant²
∴cos²x=1/(1+tan²x)=1/(1+t^4)，dx=darctant²=2t/(1+t^4) dt
∴原式=∫ 1/[1/(1+t^4)*t]*2t/(1+t^4) dt
=∫ 2 dt=2t+C=2√tanx+C

Answer 3

如图，OK

Answer 4

Answer 5

(1)
∫x.√(1-x^2) dx
=-(1/2)∫√(1-x^2) d(1-x^2)
= -(1/3) (1-x^2)^(3/2) +C
(2)
let
u=√tanx
du = { (secx)^2/[2√tanx] } dx
2u du = dx/(cosx)^2
∫ dx/[ (cosx)^2.√tanx ]

=∫ 2udu/u
=2∫ du
=2u + C
=2√tanx + C