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Popular Calculus >

integral of (sin(npix))/(npi)

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Solution

∫nπsin(nπx)​dx

Solution

−π2n21​cos(πnx)+C
Solution steps
∫nπsin(nπx)​dx
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=nπ1​⋅∫sin(nπx)dx
Apply u-substitution
=nπ1​⋅∫πnsin(u)​du
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=nπ1​⋅πn1​⋅∫sin(u)du
Use the common integral: ∫sin(u)du=−cos(u)=nπ1​⋅πn1​(−cos(u))
Substitute back u=nπx=nπ1​⋅πn1​(−cos(nπx))
Simplify nπ1​⋅πn1​(−cos(nπx)):−π2n21​cos(πnx)
=−π2n21​cos(πnx)
Add a constant to the solution=−π2n21​cos(πnx)+C

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Frequently Asked Questions (FAQ)

  • What is the integral of (sin(npix))/(npi) ?

    The integral of (sin(npix))/(npi) is -1/(pi^2n^2)cos(pinx)+C
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