第1个回答 2019-09-15
=∫2/(2-sin²x)-1dx
=√2/2∫1/(√2-sinx)+1/(√2+sinx)dx-x
其中
∫1/(√2-sinx)dx
=∫1/(√2+1-2cos²(x/2-π/4)dx
=∫sec²(x/2-π/4)/((√2+1)sec²(x/2-π/4)-2)dx
=2∫1/(√2+1)tan²(x/2-π/4)+(√2-1))dtan(x/2-π/4)
=2arctan((√2+1)tan(x/2-π/4))
同理
∫1/(√2+sinx)dx=2arctan((√2-1)tan(x/2-π/4))
∴不定积分
=√2(arctan((√2+1)tan(x/2-π/4))+arctan((√2-1)tan(x/2-π/4)))-x+C本回答被提问者和网友采纳