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Popular Calculus >

integral of xsin((npix)/2)

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Solution

∫xsin(2nπx​)dx

Solution

π2n24​(−2πn​xcos(2πnx​)+sin(2πnx​))+C
Solution steps
∫xsin(2nπx​)dx
Apply u-substitution
=∫π2n24usin(u)​du
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=π2n24​⋅∫usin(u)du
Apply Integration By Parts
=π2n24​(−ucos(u)−∫−cos(u)du)
∫−cos(u)du=−sin(u)
=π2n24​(−ucos(u)−(−sin(u)))
Substitute back u=2nxπ​=π2n24​(−2nxπ​cos(2nxπ​)−(−sin(2nxπ​)))
Simplify π2n24​(−2nxπ​cos(2nxπ​)−(−sin(2nxπ​))):π2n24​(−2πn​xcos(2πnx​)+sin(2πnx​))
=π2n24​(−2πn​xcos(2πnx​)+sin(2πnx​))
Add a constant to the solution=π2n24​(−2πn​xcos(2πnx​)+sin(2πnx​))+C

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Frequently Asked Questions (FAQ)

  • What is the integral of xsin((npix)/2) ?

    The integral of xsin((npix)/2) is 4/(pi^2n^2)(-(pin)/2 xcos((pinx)/2)+sin((pinx)/2))+C
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