Integral of 5sin(x/2+п/4) 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:
∫5sin(2x+4π)dx=5∫sin(2x+4π)dx
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Let u=2x+4π.
Then let du=2dx and substitute 2du:
∫4sin(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫2sin(u)du=2∫sin(u)du
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The integral of sine is negative cosine:
∫sin(u)du=−cos(u)
So, the result is: −2cos(u)
Now substitute u back in:
−2cos(2x+4π)
So, the result is: −10cos(2x+4π)
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Now simplify:
−10cos(2x+4π)
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Add the constant of integration:
−10cos(2x+4π)+constant
The answer is:
−10cos(2x+4π)+constant
The answer (Indefinite)
[src]
/
|
| /x pi\ /pi x\
| 5*sin|- + --| dx = C - 10*cos|-- + -|
| \2 4 / \4 2/
|
/
−10cos(2x+4π)
The graph
5(2cos(2π)−2cos(43π))
=
Use the examples entering the upper and lower limits of integration.