高等数学，求定积分，最好有解答过程！谢谢！

Answer 1

(1)
令√x=t，则x=t²
x：0→4，t：0→2
∫[0：4][1/(1+√x)]dx
=∫[0：2][1/(1+t)]d(t²)
=∫[0：2][2t/(1+t)]dt
=∫[0：2][(2t+2-2)/(1+t)]dt
=2∫[0：2]dt -2∫[0：2][1/(1+t)]d(1+t)
=2t|[0：2]-2ln|1+t||[0：2]
=2·(2-0)-2(ln3-ln1)
=4-2ln3
(2)
令√(u-1)=t，则u=t²+1
u：1→5，t：0→2
∫[1：5][√(u-1)/u]du
=∫[0：2][t/(1+t²)d(1+t²)
=∫[0：2][2t²/(1+t²)dt
=∫[0：2][(2t²+2-2)/(1+t²)dt
=∫[0：2]2dt -2∫[0：2][1/(1+t²)]dt
=2t|[0：2]-2arctant|[0：2]
=2·(2-0)-2(arctan2-arctan0)
=4-2arctan2
(3)
∫[0：π]x²cos(2x)dx
=½∫[0：π]x²d[sin(2x)]
=½x²·sin(2x)|[0：π]-½∫[0：π]sin(2x)d(x²)
=½·[π²·sin(2π)-0²·sin0]-½∫[0：π]2xsin(2x)dx
=½∫[0：π]xd[cos(2x)]
=½x·cos(2x)|[0：π] -½∫[0：π]cos(2x)dx
=½·[π·cos(2π)-0·cos0]-¼sin(2x)|[0：π]
=½π-¼[sin(2π)-sin0]
=½π

Answer 2

设 x = t²。则 dx =2t*dt，t 的积分范围变换为：[0, 2] 

那么，上式积分变换为：

=∫2t*dt/(1+t)

=2∫[1-1/(1+t)]*dt

=2[∫dt - ∫dt/(1+t)]

=2[t - ln(1+t)]|t=0→2

=2[(2-0) - (ln3-ln1)]

=2(2-ln3)

=4-2ln3

设 u - 1=t²。则 du = 2t*dt，t 的积分范围变换为：[0,2]

那么，上式积分变换为：

=∫t/(1+t²)*2t*dt

=2∫t²*dt/(1+t²)

=2∫[1-dt/(1+t²)]

=2[∫dt - ∫dt/(1+t²)]

=2[t - arctan(t)]|t=0→2

=2[(2-0) - (arctan2 - arctan0)]

=4 - 2arctan2

∫x²cos2x*dx

=1/2*x²*sin2x - 1/2*∫sin2x * 2x*dx

=1/2*x²*sin2x - ∫x*sin2x*dx

=1/2*x²*sin2x - 1/2*x*sin2x + 1/2*∫sin2x*dx

=[1/2*x²*sin2x - 1/2*x*sin2x -1/4*cos2x]|x=0→π

=1/2*[π²*sin(2π)-0²*sin0] - 1/2*[π*sin(2π) - 0*sin0] -1/4*[cos2π - cos0]

=0