Question

求高数大神解一下这四道题，最好要完整的步骤…💗

Answer 1

我也想看看啊

Answer 2

(1)原式=∫1/√(x-x²)dx

=∫1/√[(1/4)-(x-1/2)²]dx

=∫2/√[1-(2x-1)²]dx

=∫1/√[1-(2x-1)²]d(2x-1)

=arcsin(2x-1)+C

(2)原式=∫(1+sin²x)/[1+(2cos²x-1)]dx

=∫(1+sin²x)/(2cos²x)dx

=(1/2)∫[(1/cos²x)+(sin²x/cos²x)]dx

=(1/2)∫(sec²x+tan²x)dx

=(1/2)∫[sec²x+(sec²x-1)]dx

=(1/2)∫(2sec²x-1)dx

=(1/2)·(2tanx-x)+C

(3)∫e^x·cos2xdx

=(1/2)∫e^xd(sin2x)

=(1/2)[e^x·sin2x-∫sin2x·e^xdx]

=(1/2)e^x·sin2x+(1/4)∫e^xd(cos2x)

=(1/2)e^x·sin2x+(1/4)[e^x·cos2x-∫cos2x·e^x]dx

=(1/2)e^x·sin2x+(1/4)e^x·cos2x-(1/4)∫e^x·cos2xdx

==> (5/4)∫e^x·cos2xdx=(1/2)e^x·sin2x+(1/4)e^x·cos2x

==> ∫e^x·cos2xdx=(2/5)e^x·sin2x+(1/5)e^x·cos2x+C

(4)∫<0,1>[x/(1+x²)²]dx

=(1/2)∫<0,1>[1/(1+x²)²]d(1+x²)

=(-1/2)[1/(1+x²)]|<0,1>

=(-1/2)[(1/2)-1]

=1/4

Answer 3

希望有所帮助