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Popular Calculus >

integral from 0 to pi/2 of x^2ycos(xy^2)

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Solution

∫02π​​x2ycos(xy2)dx

Solution

4y5π2y4sin(2πy2​)−8sin(2πy2​)+4πy2cos(2πy2​)​
Solution steps
∫02π​​x2ycos(xy2)dx
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=y⋅∫02π​​cos(y2x)x2dx
Simplify y2:yy
=y⋅∫02π​​cos(yyx)x2dx
Apply u-substitution
=y⋅∫02πy2​​y6u2cos(u)​du
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=yy61​⋅∫02πy2​​u2cos(u)du
Apply Integration By Parts
=yy61​[u2sin(u)−∫2usin(u)du]02πy2​​
∫2usin(u)du=2(−ucos(u)+sin(u))
=yy61​[u2sin(u)−2(−ucos(u)+sin(u))]02πy2​​
Simplify yy61​[u2sin(u)−2(−ucos(u)+sin(u))]02πy2​​:y51​[u2sin(u)−2(−ucos(u)+sin(u))]02πy2​​
=y51​[u2sin(u)−2(−ucos(u)+sin(u))]02πy2​​
Compute the boundaries:4π2y4​sin(2πy2​)−2sin(2πy2​)+πy2cos(2πy2​)
=y51​(4π2y4​sin(2πy2​)−2sin(2πy2​)+πy2cos(2πy2​))
Simplify=4y5π2y4sin(2πy2​)−8sin(2πy2​)+4πy2cos(2πy2​)​

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