To evaluate the integral problem: `int_0^(pi/2) xsin(2x) dx` ,we may first solve for its indefinite integral. Indefinite integral are written in the form of `int f(x) dx = F(x) +C`
where: `f(x)` as the integrand
`F(x)` as the anti-derivative function
`C` as the arbitrary constant known as constant of integration
We follow a formula from basic integration table to determine the indefinite integral function `F(x)`...
To evaluate the integral problem: `int_0^(pi/2) xsin(2x) dx` ,we may first solve for its indefinite integral. Indefinite integral are written in the form of `int f(x) dx = F(x) +C`
where: `f(x)` as the integrand
`F(x)` as the anti-derivative function
`C` as the arbitrary constant known as constant of integration
We follow a formula from basic integration table to determine the indefinite integral function `F(x)` . For the integrals with logarithm, the problem resembles the formula:
`int x sin(ax) dx= -(xcos(ax))/a+sin(ax)/a^2 +C` .
By comparing `x sin(ax) ` with` xsin(2x)` , we determine that `a= 2` .
Plug-in `a=2` to the integral formula, we get:
`int_0^(pi/2) xsin(2x) dx=-(xcos((2)x))/(2)+sin((2)x)/(2)^2|_0^(pi/2)`
`=-(xcos(2x))/2+sin(2x)/4|_0^(pi/2)`
After solving the indefinite integral from, we may apply definite integral formula:` F(x)|_a^b = F(b) - F(a)` .
`-(xcos(2x))/2+sin(2x)/4|_0^(pi/2) =[-((pi/2) *cos(2*(pi/2)))/2+sin(2*(pi/2))/4]-[-(0*cos(2*0))/2+sin(2*0)/4 ]`
`=[-((pi/2) *cos(pi ))/2+sin(pi) /4]-[-(0*cos(0))/2+sin(0)/4 ]`
`=[-(pi*(-1))/4+0 /4]-[-(0*1)/2+(0)/4 ]`
`=[pi/4+0]-[0+0]`
`= [pi/4] - [0]`
`=pi/4`
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