Question
Evaluate the integral
21xsin(2x)+41cos(2x)+C,C∈R
Evaluate
∫xcos(2x)dx
Prepare for integration by parts
u=xdv=cos(2x)dx
Calculate the derivative
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Calculate the derivative
u=x
Evaluate the derivative
du=x′dx
Evaluate the derivative
du=1dx
Simplify the expression
du=dx
du=dxdv=cos(2x)dx
Evaluate the integral
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Evaluate the integral
dv=cos(2x)dx
Evaluate the integral
∫1dv=∫cos(2x)dx
Evaluate the integral
v=21sin(2x)
du=dxv=21sin(2x)
Substitute u=x、v=21sin(2x)、du=dx、dv=cos(2x)dx for ∫udv=uv−∫vdu
x×21sin(2x)−∫1×21sin(2x)dx
Calculate
21xsin(2x)−∫21sin(2x)dx
Evaluate the integral
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Evaluate the integral
−∫21sin(2x)dx
Use the property of integral ∫kf(x)dx=k∫f(x)dx
−21×∫sin(2x)dx
Use the property of integral ∫sin(ax)dx=−a1cos(ax)
−21(−21)cos(2x)
Calculate
−21(−21cos(2x))
Rewrite the expression
21×21cos(2x)
Multiply the terms
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Evaluate
21×21
To multiply the fractions,multiply the numerators and denominators separately
2×21
Multiply the numbers
41
41cos(2x)
21xsin(2x)+41cos(2x)
Solution
21xsin(2x)+41cos(2x)+C,C∈R
Show Solution
