∫1/((cosx)^4+(sinx)^4)dx求∫1/((cosx)^4+(sinx)^4)dx

Question

∫1/((cosx)^4+(sinx)^4)dx
求∫1/((cosx)^4+(sinx)^4)dx

Answer 1

(cosx)^4+(sinx)^4
=（cos² x+sin² x）-2cos² xsin² x
=1-1/2sin² 2x
=cos² 2x+sin² 2x-1/2sin² 2x
=cos² 2x+1/2sin² 2x
∫1/((cosx)^4+(sinx)^4)
=∫1/（cos² 2x+1/2sin² 2x）dx
=∫sec² 2x/（1+1/2tan² 2x）dx
=∫1/2*1/(1+1/2tan² 2x）d(tan2x)
=∫1/(2+tan² 2x）d(tan2x) (公式∫1/(a^2+x^2)dx=1/a*arctan(x/a)+c )
=（√2）/2arctan[√2(tan 2x）/2]+C