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Popular Calculus >

integral from 0 to pi of 3sin^2(xco)s^4x

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Solution

∫0π​3sin2(xco)s4xdx

Solution

8c2o23s4(2π2c2o2−2πcosin(2πco)−cos(2πco)+1)​
Solution steps
∫0π​3sin2(xco)s4xdx
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=3s4⋅∫0π​sin2(cox)xdx
Apply u-substitution
=3s4⋅∫0coπ​c2o2usin2(u)​du
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=3s4c2o21​⋅∫0coπ​usin2(u)du
Apply Integration By Parts
=3s4c2o21​[21​u(u−21​sin(2u))−∫21​(u−21​sin(2u))du]0πco​
∫21​(u−21​sin(2u))du=21​(2u2​+41​cos(2u))
=3s4c2o21​[21​u(u−21​sin(2u))−21​(2u2​+41​cos(2u))]0πco​
Simplify 3s4c2o21​[21​u(u−21​sin(2u))−21​(2u2​+41​cos(2u))]0πco​:c2o23s4​[82u2−2usin(2u)−cos(2u)​]0πco​
=c2o23s4​[82u2−2usin(2u)−cos(2u)​]0πco​
Compute the boundaries:82π2c2o2−2πcosin(2πco)−cos(2πco)​+81​
=c2o23s4​(82π2c2o2−2πcosin(2πco)−cos(2πco)​+81​)
Simplify=8c2o23s4(2π2c2o2−2πcosin(2πco)−cos(2πco)+1)​

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