Integral of xcos(pix) dx
The solution
Detail solution
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Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=x and let dv(x)=cos(πx).
Then du(x)=1.
To find v(x):
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Let u=πx.
Then let du=πdx and substitute πdu:
∫πcos(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫cos(u)du=π∫cos(u)du
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The integral of cosine is sine:
∫cos(u)du=sin(u)
So, the result is: πsin(u)
Now substitute u back in:
πsin(πx)
Now evaluate the sub-integral.
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The integral of a constant times a function is the constant times the integral of the function:
∫πsin(πx)dx=π∫sin(πx)dx
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Let u=πx.
Then let du=πdx and substitute πdu:
∫πsin(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫sin(u)du=π∫sin(u)du
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The integral of sine is negative cosine:
∫sin(u)du=−cos(u)
So, the result is: −πcos(u)
Now substitute u back in:
−πcos(πx)
So, the result is: −π2cos(πx)
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Now simplify:
π2πxsin(πx)+cos(πx)
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Add the constant of integration:
π2πxsin(πx)+cos(πx)+constant
The answer is:
π2πxsin(πx)+cos(πx)+constant
The answer (Indefinite)
[src]
/
| cos(pi*x) x*sin(pi*x)
| x*cos(pi*x) dx = C + --------- + -----------
| 2 pi
/ pi
∫xcos(πx)dx=C+πxsin(πx)+π2cos(πx)
The graph
−π22
=
−π22
Use the examples entering the upper and lower limits of integration.