Question

高数题2道。求解！！！！！！！！

Answer 1

1、=cos(pi*cos^2x)(-sinx)-cos(pi*sin^2x)*cosx
=cos(pi*sin^2x)sinx--cos(pi*sin^2x)*cosx
=cos(pi*sin^2x)(sinx--cosx)。
2、=积分（从0到pi/4）xdx/(2cos^2x)
=0.5积分（从0到pi/4）xdtanx

=0.5xtanx|上限pi/4下限0--0.5积分（从0到pi/4）tanxdx
=pi/8+0.5lncosx|上限pi/4下限0
=pi/8--0.5ln根号(2)
=pi/8--(ln2)/4

Answer 2

1.∫(sinx,cosx)cos(πt^2)dx
=∫(sinx,0)cos(πt^2)dx+∫(0,cosx)cos(πt^2)dx
=-∫(0,sinx)cos(πt^2)dx+∫(0,cosx)cos(πt^2)dx
所以：（∫(sinx,cosx)cos(πt^2)dx)'
=-cos(π(sinx)^2)*cosx-cos(π(cosx)^2)*sinx
2.下午来