Integral of sinxlncosx dx
The solution
Detail solution
-
There are multiple ways to do this integral.
Method #1
-
Let u=log(cos(x)).
Then let du=−cos(x)sin(x)dx and substitute −du:
∫ueudu
-
The integral of a constant times a function is the constant times the integral of the function:
∫(−ueu)du=−∫ueudu
-
Use integration by parts:
∫udv=uv−∫vdu
Let u(u)=u and let dv(u)=eu.
Then du(u)=1.
To find v(u):
-
The integral of the exponential function is itself.
∫eudu=eu
Now evaluate the sub-integral.
-
The integral of the exponential function is itself.
∫eudu=eu
So, the result is: −ueu+eu
Now substitute u back in:
−log(cos(x))cos(x)+cos(x)
Method #2
-
Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=log(cos(x)) and let dv(x)=sin(x).
Then du(x)=−cos(x)sin(x).
To find v(x):
-
The integral of sine is negative cosine:
∫sin(x)dx=−cos(x)
Now evaluate the sub-integral.
-
The integral of sine is negative cosine:
∫sin(x)dx=−cos(x)
-
Now simplify:
(1−log(cos(x)))cos(x)
-
Add the constant of integration:
(1−log(cos(x)))cos(x)+constant
The answer is:
(1−log(cos(x)))cos(x)+constant
The answer (Indefinite)
[src]
/
|
| sin(x)*log(cos(x)) dx = C - cos(x)*log(cos(x)) + cos(x)
|
/
cosx−cosxlogcosx
The graph
-1 - cos(1)*log(cos(1)) + cos(1)
−cos1logcos1+cos1−1
=
-1 - cos(1)*log(cos(1)) + cos(1)
−1−log(cos(1))cos(1)+cos(1)
Use the examples entering the upper and lower limits of integration.