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,
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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,
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,
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,
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