Question

不定积分求解。

Answer 1

详解见图片！

Answer 2

=(1/2)∫[2sinxcosx-1+1]/(sinx+cosx) dx
=(1/2)∫(2sinxcosx+sin²x+cos²x)/(sinx+cosx) dx － (1/2)∫1/(sinx+cosx)dx
=(1/2)∫(sinx+cosx)² / (sinx+cosx) dx - (1/2)∫1 / [ (√2)sin(x+ π/4)] dx
=(1/2)∫(sinx+cosx) dx - (√2 / 4)∫csc(x+ π/4) d (x + π/4)
=(1/2)( sinx - cosx) - (√2 / 4) ln| csc(x+ π/4) - cot(x + π/4) | + C
=(1/2)( sinx - cosx) - (√2 / 4) ln| tan(x/2 + π/8) | + C

Answer 3

=∫(sinx-cosx)^2/(sin^2x-cos^2x)dx
=∫(1-sin2x)/(-cos2x)dx
=-∫(1/cos2x)dx-1/2∫1/cos2xd(cos2x)
=-1/2∫1/cos^2(2x)d(sin2x)-1/2ln|cos2x|
=-1/4∫[(1/(1-sin2x)+(1/(1+sin2x)]d(sin2x)-1/2ln|cos2x|
=-1/4[-ln(1-sin2x)+ln(1+sin2x)]-1/2ln|cos2x|+C
=-ln|sinx+cosx|+C