1/sin2x的原函数是?要求有解答过程!在线等!
2019-05-28
1/sin2x的原函数是?要求有解答过程!在线等!
优质解答
∫dx/sin2x
=∫sin2xdx/(sin2x)^2
=-1/2*∫d(cos2x)/[1-(cos2x)^2]
=-1/4*∫d(cos2x)/[1/(1+cos2x)+1/(1-cos2x)]
=-1/4*[ln(1+cos2x)-ln(1-cos2x)]+C
=-1/4*ln[(1+cos2x)/(1-cos2x)]+C
∫dx/sin2x
=∫sin2xdx/(sin2x)^2
=-1/2*∫d(cos2x)/[1-(cos2x)^2]
=-1/4*∫d(cos2x)/[1/(1+cos2x)+1/(1-cos2x)]
=-1/4*[ln(1+cos2x)-ln(1-cos2x)]+C
=-1/4*ln[(1+cos2x)/(1-cos2x)]+C