一道积分问题：∫sinh((lnx))sin((lnx))dx

Question

一道积分问题：∫sinh((lnx))sin((lnx))dx

如果不方便可以写在纸上再上传图片，一样可以接受，谢谢了。

Answer 1

Q: ∫sinh((lnx))sin((lnx)) dx

Let u = ln(x)
∴ du = (1/x) dx

Note that e^(u)sinh((lnx))
= e^(u) [ e^ln(x) - e^-ln(x) ] / 2
= e^(u)(e^(u) - e^(-u)) /2
= (1/2) (e^(2u) -1)

∫ sinh((lnx))sin((lnx)) dx
= ∫ e^(u)sin(u)sinh(u)du (add in e^u, so you can utilize above formula)
= ∫ (1/2)(e^(2u)-1)sin(u) du (substitute in above formula)
= 1/2 ∫ (e^(2u)sin(u) - sin(u))du
= 1/2 ∫ e^(2u)sin(u)du - 1/2 ∫ sin(u)du

You have two choices here:

1) Use integration by parts to solve the first integral, ∫ e^(2u)sin(u)du
2) Use the formula, ∫ e^(αu)sin(βu)du = e^(αu)(-β cos(βu) + αsin(βu)) / (α²+ β²)、

I chose the latter one.

= 1/5 e^(2u)sin(u) - 1/10 e^(2u)cos(u) - 1/2 ∫ sin(u) u
= 1/5 e^(2u)sin(u) - 1/10 e^(2u)cos(u) + cos(u)/2 + C

Now substitude u = ln(x) back into the equation.

= 1/5 x²sin(ln(x)) - 1/10 x²cos(ln(x)) + 1/2 cos(ln(x)) + C
= (1/10)(2x²sin(ln(x)) - (x²-5)cos(ln(x)) + C

∴ (1/10)(2x²sin(ln(x)) - (x²-5)cos(ln(x)) + C

Answer 2

∫sin(lnx)dx =xsin(lnx)-∫xdsin(lnx) =xsin(lnx)-∫x*cos(lnx)*1/xdx =xsin(lnx)-∫cos(lnx)dx =xsin(lnx)-xcos(lnx)+∫xdcos(lnx)