Integral of 1/(2sinx+sin2x) dx
The solution
The answer (Indefinite)
[src]
−8sin2(2x)+32sinxsin(2x)+8cos2(2x)+(32cosx+16)cos(2x)+32sin2x+32cos2x+32cosx+8(sin2(2x)+4sinxsin(2x)+cos2(2x)+(4cosx+2)cos(2x)+4sin2x+4cos2x+4cosx+1)log(sin2x+cos2x+2cosx+1)+(−sin2(2x)−4sinxsin(2x)−cos2(2x)+(−4cosx−2)cos(2x)−4sin2x−4cos2x−4cosx−1)log(sin2x+cos2x−2cosx+1)−4sinxsin(2x)−4cosxcos(2x)−8sin2x−8cos2x−4cosx
1
/
|
| 1
| ------------------- dx
| 2*sin(x) + sin(2*x)
|
/
0
=
1
/
|
| 1
| ------------------- dx
| 2*sin(x) + sin(2*x)
|
/
0
0∫12sin(x)+sin(2x)1dx
Use the examples entering the upper and lower limits of integration.