Question

不定积分sinx/(1+sinx+cosx)

不定积分sinx/(1+sinx+cosx)不定积分sinx/(1+sinx+cosx)

Answer 1

∫ sinx/(1+sinx+cosx) dx

=∫sinx(sinx+cosx-1)/[(sinx+cosx+1)(sinx+cosx-1)] dx

=∫(sin^2x+sinxcosx-sinx)/[(sinx+cosx)^2-1] dx

=∫(sin^2x+sinxcosx-sinx)/(2sinxcosx) dx

=(1/2)∫sinx/cosx dx+(1/2)∫ dx-(1/2)∫1/cosx dx

=(-1/2)∫1/cosx d(cosx)+(1/2)∫ dx-(1/2)∫secx dx

=(1/2)x-(1/2)ln(cosx)+(1/2)x-(1/2)ln(secx+tanx)+C

扩展资料

sin30° = 1 / 2

sin45° = √2 / 2

sin60° = √3 / 2

tan30° = 1 / √3 = √3 / 3

tan45° = 1

tan60° = √3

正弦函数sin（A）=a/c

余弦函数cos（A）=b/c

正切函数tan（A）=a/b

余切函数cot（A）=b/a

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

Answer 2

简单计算一下即可，答案如图所示

Answer 3

∫ sinx/(1+sinx+cosx) dx
=∫sinx(sinx+cosx-1)/[(sinx+cosx+1)(sinx+cosx-1)] dx
=∫(sin^2x+sinxcosx-sinx)/[(sinx+cosx)^2-1] dx
=∫(sin^2x+sinxcosx-sinx)/(2sinxcosx) dx
=(1/2)∫sinx/cosx dx+(1/2)∫ dx-(1/2)∫1/cosx dx
=(-1/2)∫1/cosx d(cosx)+(1/2)∫ dx-(1/2)∫secx dx
=(1/2)x-(1/2)ln(cosx)+(1/2)x-(1/2)ln(secx+tanx)+C