Integral of 2^(x*(-3))*sin(2*x) dx
The solution
Detail solution
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The integral of a constant times a function is the constant times the integral of the function:
∫2⋅2(−3)xsin(x)cos(x)dx=2∫2(−3)xsin(x)cos(x)dx
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Don't know the steps in finding this integral.
But the integral is
4⋅23x+9⋅23xlog(2)2sin2(x)−4⋅23x+9⋅23xlog(2)23log(2)sin(x)cos(x)−4⋅23x+9⋅23xlog(2)2cos2(x)
So, the result is: 4⋅23x+9⋅23xlog(2)22sin2(x)−4⋅23x+9⋅23xlog(2)26log(2)sin(x)cos(x)−4⋅23x+9⋅23xlog(2)22cos2(x)
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Now simplify:
−4+9log(2)28−x(log(8)sin(2x)+2cos(2x))
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Add the constant of integration:
−4+9log(2)28−x(log(8)sin(2x)+2cos(2x))+constant
The answer is:
−4+9log(2)28−x(log(8)sin(2x)+2cos(2x))+constant
The answer (Indefinite)
[src]
/
| 2 2
| x*(-3) 2*cos (x) 2*sin (x) 6*cos(x)*log(2)*sin(x)
| 2 *sin(2*x) dx = C - ----------------------- + ----------------------- - -----------------------
| 3*x 3*x 2 3*x 3*x 2 3*x 3*x 2
/ 4*2 + 9*2 *log (2) 4*2 + 9*2 *log (2) 4*2 + 9*2 *log (2)
∫2(−3)xsin(2x)dx=C+4⋅23x+9⋅23xlog(2)22sin2(x)−4⋅23x+9⋅23xlog(2)26log(2)sin(x)cos(x)−4⋅23x+9⋅23xlog(2)22cos2(x)
The graph
2
-------------
2
4 + 9*log (2)
4+9log(2)22
=
2
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2
4 + 9*log (2)
4+9log(2)22
Use the examples entering the upper and lower limits of integration.