Integral of xcos(2x)x \cos{\left(2 x \right)}

The calculator will find the integral/antiderivative of xcos(2x)x \cos{\left(2 x \right)}, with steps shown.

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Your Input

Find xcos(2x)dx\int x \cos{\left(2 x \right)}\, dx.

Solution

For the integral xcos(2x)dx\int{x \cos{\left(2 x \right)} d x}, use integration by parts udv=uvvdu\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}.

Let u=x\operatorname{u}=x and dv=cos(2x)dx\operatorname{dv}=\cos{\left(2 x \right)} dx.

Then du=(x)dx=1dx\operatorname{du}=\left(x\right)^{\prime }dx=1 dx (steps can be seen ») and v=cos(2x)dx=sin(2x)2\operatorname{v}=\int{\cos{\left(2 x \right)} d x}=\frac{\sin{\left(2 x \right)}}{2} (steps can be seen »).

The integral becomes

xcos(2x)dx=(xsin(2x)2sin(2x)21dx)=(xsin(2x)2sin(2x)2dx){\color{red}{\int{x \cos{\left(2 x \right)} d x}}}={\color{red}{\left(x \cdot \frac{\sin{\left(2 x \right)}}{2}-\int{\frac{\sin{\left(2 x \right)}}{2} \cdot 1 d x}\right)}}={\color{red}{\left(\frac{x \sin{\left(2 x \right)}}{2} - \int{\frac{\sin{\left(2 x \right)}}{2} d x}\right)}}

Apply the constant multiple rule cf(x)dx=cf(x)dx\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx with c=12c=\frac{1}{2} and f(x)=sin(2x)f{\left(x \right)} = \sin{\left(2 x \right)}:

xsin(2x)2sin(2x)2dx=xsin(2x)2(sin(2x)dx2)\frac{x \sin{\left(2 x \right)}}{2} - {\color{red}{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}} = \frac{x \sin{\left(2 x \right)}}{2} - {\color{red}{\left(\frac{\int{\sin{\left(2 x \right)} d x}}{2}\right)}}

Let u=2xu=2 x.

Then du=(2x)dx=2dxdu=\left(2 x\right)^{\prime }dx = 2 dx (steps can be seen »), and we have that dx=du2dx = \frac{du}{2}.

Therefore,

xsin(2x)2sin(2x)dx2=xsin(2x)2sin(u)2du2\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{2} = \frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{2}

Apply the constant multiple rule cf(u)du=cf(u)du\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du with c=12c=\frac{1}{2} and f(u)=sin(u)f{\left(u \right)} = \sin{\left(u \right)}:

xsin(2x)2sin(u)2du2=xsin(2x)2(sin(u)du2)2\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{2} = \frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{2}

The integral of the sine is sin(u)du=cos(u)\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}:

xsin(2x)2sin(u)du4=xsin(2x)2(cos(u))4\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{4} = \frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{4}

Recall that u=2xu=2 x:

xsin(2x)2+cos(u)4=xsin(2x)2+cos((2x))4\frac{x \sin{\left(2 x \right)}}{2} + \frac{\cos{\left({\color{red}{u}} \right)}}{4} = \frac{x \sin{\left(2 x \right)}}{2} + \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{4}

Therefore,

xcos(2x)dx=xsin(2x)2+cos(2x)4\int{x \cos{\left(2 x \right)} d x} = \frac{x \sin{\left(2 x \right)}}{2} + \frac{\cos{\left(2 x \right)}}{4}

Add the constant of integration:

xcos(2x)dx=xsin(2x)2+cos(2x)4+C\int{x \cos{\left(2 x \right)} d x} = \frac{x \sin{\left(2 x \right)}}{2} + \frac{\cos{\left(2 x \right)}}{4}+C

Answer

xcos(2x)dx=(xsin(2x)2+cos(2x)4)+C\int x \cos{\left(2 x \right)}\, dx = \left(\frac{x \sin{\left(2 x \right)}}{2} + \frac{\cos{\left(2 x \right)}}{4}\right) + CA