Using integration by parts, how would I solve the integral of (x sin(4x))dx evaluated from pi/2 to 0?
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Answer:
Using two steps of U*dV substitution (with some built in U substitution), you get: (sin(4x))/16-(x*cos(4x))/4 Evaluated at zero gives you nothing since sin(0)=0 and there is an x in the second fraction. Now all you have to do is evaluate at pi/2. (sin(2pi))/16-(pi/2*cos(2pi))/4 -pi/2/4 = -pi/8
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Other answers
let u = x so du = dx and dv = sin(4x)dx then v = -cos(4x)/4 , since dv/dx = 4 sin(4x)/4 = sin(4x) So integrating by parts we have: integral udv = uv - integral vdu So integral xsin(4x)dx= -xcos(4x)/4 - integral -(1/4)cos(4x)dx = -xcos(4x)/4 + sine(4x)/4 = 0.25(xcox(4x)+sine(4x)) Now just evaluate between pi/2 and 0.
Jimbo
=2pi
tabachoi2001
the int by parts step gives: {-(x/4)cos(4x)|pi/2 to 0}+(1/4)int cos(4x)dx of the same limits. ={0+pi/8}+(1/16)sin(4x)|pi/2 to 0 =pi/8+0-0 =pi/8
Paul C
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