Integral of cos(lnx)/x^2 dx
The solution
Detail solution
-
Let u=log(x).
Then let du=xdx and substitute du:
∫e−ucos(u)du
-
There are multiple ways to do this integral.
Method #1
-
Let u=−u.
Then let du=−du and substitute −du:
∫(−eucos(u))du
-
The integral of a constant times a function is the constant times the integral of the function:
∫eucos(u)du=−∫eucos(u)du
-
Use integration by parts, noting that the integrand eventually repeats itself.
-
For the integrand eucos(u):
Let u(u)=cos(u) and let dv(u)=eu.
Then ∫eucos(u)du=eucos(u)−∫(−eusin(u))du.
-
For the integrand −eusin(u):
Let u(u)=−sin(u) and let dv(u)=eu.
Then ∫eucos(u)du=eusin(u)+eucos(u)+∫(−eucos(u))du.
-
Notice that the integrand has repeated itself, so move it to one side:
2∫eucos(u)du=eusin(u)+eucos(u)
Therefore,
∫eucos(u)du=2eusin(u)+2eucos(u)
So, the result is: −2eusin(u)−2eucos(u)
Now substitute u back in:
2e−usin(u)−2e−ucos(u)
Method #2
-
Use integration by parts, noting that the integrand eventually repeats itself.
-
For the integrand e−ucos(u):
Let u(u)=cos(u) and let dv(u)=e−u.
Then ∫e−ucos(u)du=−∫e−usin(u)du−e−ucos(u).
-
For the integrand e−usin(u):
Let u(u)=sin(u) and let dv(u)=e−u.
Then ∫e−ucos(u)du=∫(−e−ucos(u))du+e−usin(u)−e−ucos(u).
-
Notice that the integrand has repeated itself, so move it to one side:
2∫e−ucos(u)du=e−usin(u)−e−ucos(u)
Therefore,
∫e−ucos(u)du=2e−usin(u)−2e−ucos(u)
Now substitute u back in:
2xsin(log(x))−2xcos(log(x))
-
Now simplify:
−2x2cos(log(x)+4π)
-
Add the constant of integration:
−2x2cos(log(x)+4π)+constant
The answer is:
−2x2cos(log(x)+4π)+constant
The answer (Indefinite)
[src]
/
|
| cos(log(x)) sin(log(x)) cos(log(x))
| ----------- dx = C + ----------- - -----------
| 2 2*x 2*x
| x
|
/
∫x2cos(log(x))dx=C+2xsin(log(x))−2xcos(log(x))
⟨−∞,∞⟩
=
⟨−∞,∞⟩
Use the examples entering the upper and lower limits of integration.