Question

∫sinxcosx/(1+sin^4x)dx

求不定积分,要过程,谢谢啊!

Answer 1

∫sinxcosx/(1+sin^4x)dx
=∫sinx/(1+sin^4x)d(sinx)
=1/2*∫1/(1+(sin^2x)^2)d(sin^2x)
=1/2*arctan(sin^2x)+C

Answer 2

令sinx=t
化简得
=∫t/(1+t^2)dx
=[ln(1+t^2)]/2+c
即：
ln(1+sinx^2)/2+c