How do I integrate ∫xsin(x/2)dx from 0 to pi using Integration by parts?

Answer 1

integration by parts formula: uv-∫vdu

but because we have x in the integral if we set u=x then we can use tabular integration which goes as follows

+ u=x dv=sin(x/2)dx

- u'=1dx v=-2cos(x/2)

+ u''=0 ∫v=-4sin(x/2)

the signs on the left-hand side tell us the sign of the term and we get each term by multiplying diagonally which will give us the following:

=[-2xcos(x/2) + 4sin(x/2)]₀^π

now all we have to do is use the first fundamental theorem of calculus to evaluate the actual value of the integral in our bounds.

=[-2πcos(π/2) + 4sin(π/2)] - [-2(0)cos(0/2) + 4sin(0/2)]

this then simplifies to:

= [-2π(0) + 4(1)] - [0 + 0]

which then gives us our solution:

∫₀^πxsin(x/2)dx = 4

do note that tabular integration is simply a short cut for integration by parts which we can only do when we have a polynomial term such as x or x^2 in the integral.

Answer 2

Formula integration by parts: ∫udv = uv - ∫vdu.

In our case: u = x, dv = sin(x/2)dx. Then du = dx and v = ∫sin(x/2)dx = - 2cos(x/2)∫

∫xsin(x/2)dx = - 2xcos(x/2) + ∫2cos(x/2) = - 2xscos(x/2) + 4sin(x/2) + C