Integral of exp(x)sin(x)sin(x) dx
The solution
Detail solution
-
Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=sin2(x) and let dv(x)=ex.
Then du(x)=2sin(x)cos(x).
To find v(x):
-
The integral of the exponential function is itself.
∫exdx=ex
Now evaluate the sub-integral.
-
Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=2sin(x)cos(x) and let dv(x)=ex.
Then du(x)=−2sin2(x)+2cos2(x).
To find v(x):
-
The integral of the exponential function is itself.
∫exdx=ex
Now evaluate the sub-integral.
-
Rewrite the integrand:
(−2sin2(x)+2cos2(x))ex=−2exsin2(x)+2excos2(x)
-
Integrate term-by-term:
-
The integral of a constant times a function is the constant times the integral of the function:
∫(−2exsin2(x))dx=−2∫exsin2(x)dx
-
Don't know the steps in finding this integral.
But the integral is
53exsin2(x)−52exsin(x)cos(x)+52excos2(x)
So, the result is: −56exsin2(x)+54exsin(x)cos(x)−54excos2(x)
-
The integral of a constant times a function is the constant times the integral of the function:
∫2excos2(x)dx=2∫excos2(x)dx
-
Don't know the steps in finding this integral.
But the integral is
52exsin2(x)+52exsin(x)cos(x)+53excos2(x)
So, the result is: 54exsin2(x)+54exsin(x)cos(x)+56excos2(x)
The result is: −52exsin2(x)+58exsin(x)cos(x)+52excos2(x)
-
Now simplify:
10(−2sin(2x)−cos(2x)+5)ex
-
Add the constant of integration:
10(−2sin(2x)−cos(2x)+5)ex+constant
The answer is:
10(−2sin(2x)−cos(2x)+5)ex+constant
The answer (Indefinite)
[src]
/
| 2 x 2 x x
| x 2*cos (x)*e 3*sin (x)*e 2*cos(x)*e *sin(x)
| e *sin(x)*sin(x) dx = C + ------------ + ------------ - ------------------
| 5 5 5
/
−102exsin(2x)+excos(2x)−5ex
The graph
2 2
2 2*e*cos (1) 3*e*sin (1) 2*e*cos(1)*sin(1)
- - + ----------- + ----------- - -----------------
5 5 5 5
−102esin2+ecos2−5e−52
=
2 2
2 2*e*cos (1) 3*e*sin (1) 2*e*cos(1)*sin(1)
- - + ----------- + ----------- - -----------------
5 5 5 5
−52esin(1)cos(1)−52+52ecos2(1)+53esin2(1)
Use the examples entering the upper and lower limits of integration.