Question

∫1/（2+sinx）dx

Answer 1

2+sinx=2sin(x/2)^2+2cos(x/2)^2+2sin(x/2)cos(x/2)

dx/(2+sinx)=sec(x/2)^2dx/[2+2tan(x/2)^2+2tan(x/2)]

=d(tan(x/2))/[1+tan(x/2)+tan(x/2)^2]

令u=tan(x/2)

原积分=∫du/(1+u+u^2)

=∫d(u+1/2)/[3/4+(u+1/2)^2]（用∫dx/(a^2+x^2)公式，取a=√3/2）

=1/a*arctan[(u+1/2)/a]+C

=2√3/3*arctan{[2√3tan(x/2)+√3]/3}+C

扩展资料

不定积分的公式

1、∫ a dx = ax + C，a和C都是常数

2、∫ x^a dx = [x^(a + 1)]/(a + 1) + C，其中a为常数且 a ≠ -1

3、∫ 1/x dx = ln|x| + C

4、∫ a^x dx = (1/lna)a^x + C，其中a > 0 且 a ≠ 1

5、∫ e^x dx = e^x + C

6、∫ cosx dx = sinx + C

7、∫ sinx dx = - cosx + C

8、∫ cotx dx = ln|sinx| + C = - ln|cscx| + C

9、∫ tanx dx = - ln|cosx| + C = ln|secx| + C

10、∫ secx dx =ln|cot(x/2)| + C 

= (1/2)ln|(1 + sinx)/(1 - sinx)| + C 

= - ln|secx - tanx| + C 

= ln|secx + tanx| + C

Answer 2

简单计算一下即可，答案如图所示

Answer 3

2+sinx=2sin(x/2)^2+2cos(x/2)^2+2sin(x/2)cos(x/2)
dx/(2+sinx)=sec(x/2)^2dx/[2+2tan(x/2)^2+2tan(x/2)]
=d(tan(x/2))/[1+tan(x/2)+tan(x/2)^2]
令u=tan(x/2)
原积分=∫du/(1+u+u^2)
=∫d(u+1/2)/[3/4+(u+1/2)^2]（用∫dx/(a^2+x^2)公式，取a=√3/2）
=1/a*arctan[(u+1/2)/a]+C
=2√3/3*arctan{[2√3tan(x/2)+√3]/3}+C