∫1/sin(3x)dx=? thx步骤+正确答案,谢谢

= (1/3)ln(tan(3x/2))

先令t=3x，则
∫1/sin(3x)dx
=(1/3)*∫1/sin tdt
=(1/3)*∫csc tdt
=(1/3)*(ln(csc(t)-cot(t)))
=1/3*ln(csc(3*x)-cot(3*x))
∫csc tdt很多书上都有推导过程的，这里就不详述了

∫1/sin(3x)dx
=1/3∫1/sin(3x)d3x
=1/3∫csc(3x)d3x
现在先来计算：∫cscxdx
∫cscxdx
=∫dx/sinx
=∫sinxdx/sin²x
=-∫dcosx/sin²x
=∫dcosx/(cos²x-1)
=(1/2)[∫dcosx/(cosx-1)-∫dcosx/(cosx+1)]
=(1/2)(ln|cosx-1|-ln|cosx+1|)
=(1/2)ln|(cosx-1)/(cosx+1)|
(对数里分子分母都乘以cosx-1)
=(1/2)ln|(cosx-1)²/sin²x|
=ln|(cosx-1)/sinx|
=ln|cotx-cscx|
所以
∫1/sin(3x)dx
=1/3∫1/sin(3x)d3x
=1/3∫csc(3x)d3x
=1/3ln|cot3x-csc3x|

∫1/sin(3x)dx=
设y=3x
1/3∫siny/sin^2ydy
=-1/3∫1/(1-cos^2y)dcosy
=-1/6∫[1/(1-cosy)+1/(1+cosy)]dcosy
=1/6ln...
应该不难了，
就这样吧。