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Popular Calculus >

integral from 0 to 2pi of cos^2(xsi)n^2x

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Solution

∫02π​cos2(xsi)n2xdx

Solution

8s2i2n2(8π2s2i2+4πsisin(4πsi)+cos(4πsi)−1)​
Solution steps
∫02π​cos2(xsi)n2xdx
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=n2⋅∫02π​cos2(six)xdx
Apply u-substitution
=n2⋅∫0si⋅2π​s2i2ucos2(u)​du
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=n2s2i21​⋅∫0si⋅2π​ucos2(u)du
Apply Integration By Parts
=n2s2i21​[21​u(u+21​sin(2u))−∫21​(u+21​sin(2u))du]02πsi​
∫21​(u+21​sin(2u))du=21​(2u2​−41​cos(2u))
=n2s2i21​[21​u(u+21​sin(2u))−21​(2u2​−41​cos(2u))]02πsi​
Simplify n2s2i21​[21​u(u+21​sin(2u))−21​(2u2​−41​cos(2u))]02πsi​:s2i2n2​[82u2+2usin(2u)+cos(2u)​]02πsi​
=s2i2n2​[82u2+2usin(2u)+cos(2u)​]02πsi​
Compute the boundaries:88π2s2i2+4πsisin(4πsi)+cos(4πsi)​−81​
=s2i2n2​(88π2s2i2+4πsisin(4πsi)+cos(4πsi)​−81​)
Simplify=8s2i2n2(8π2s2i2+4πsisin(4πsi)+cos(4πsi)−1)​

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