# Explain the method of integration of function y=x*sin2x?

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### 2 Answers

To integrate y = x*(sin 2x) we use integration by parts which gives us : Int [ u dv] = u*v - Int [ v du]

Let u = x , du = dx

dv = sin 2x => v = - cos 2x / 2

Int [ x*(sin 2x)]

=> x*(-cos 2x / 2) - Int [ -cos 2x / 2 dx]

=> x*(-cos 2x / 2) + sin 2x / 4

=> [sin 2x - 2x*(cos 2x)]/4

**The integral of y = x*(sin 2x) is [sin 2x - 2x*(cos 2x)]/4**

We'll use the formula sin ax = -(cos ax)'/a and we'll integrate by parts:

Int x*sin2x dx = Int x*[-(cos 2x)]' dx/2

We'll note u = x => u'du = dx

v'dv = -(cos 2x)'dx/2 => v = -(cos 2x)/2

Int udv = uv - Int vdu

Int x*sin2x dx =-x*(cos 2x)/2 - Int-(cos 2x)dx/2

Int x*sin2x dx = -x*(cos 2x)/2 + (sin 2x)/4 + C

**Therefore, using integration by parts, we've get the result: Int x*sin2x dx = -x*(cos 2x)/2 + (sin 2x)/4 + C.**