Question

一道关于不定积分的题 ∫sin平方xcosxdx/(1+sin平方x)

Answer 1

∫[(sinx)^2*cosx]/[1+(sinx)^2]dx

=∫[(sinx)^2]/[1+(sinx)^2]d(sinx)

令u=sinx

则有原式=∫u^2/(1+u^2)du=∫[1-1/(u^2+1)]du

=∫du-∫1/(u^2+1)du=u-arctanu+C=sinx-arctan(sinx)+C