Integral of x Cos2x

In this tutorial we shall find integral of x Cos2x function. To evaluate this integral we shall use the integration by parts method.

The integration of the form

I =  \int {x\cos 2xdx}


Here we have First function is x and second function will be \cos 2x

I =  \int {x\cos 2xdx} \,\,\,\,{\text{ -  -  - }}\left( {\text{i}} \right)


Using formula for integration by parts, we have

\int  {\left[ {f\left( x \right)g\left( x \right)} \right]dx = f\left( x \right)\int  {g\left( x \right)dx - \int {\left[ {\frac{d}{{dx}}f\left( x \right)\int  {g\left( x \right)dx} } \right]dx} } }


Equation (i) becomes using above formula, we have

I =  x\int {\cos 2xdx}  - \int {\left[  {\frac{d}{{dx}}x\left( {\int {\cos 2xdx} } \right)} \right]} dx


Using the integral rule \int  {\cos kx} dx = \frac{{\sin kx}}{k} + c, we have

\begin{gathered} I = x\frac{{\sin 2x}}{2} - \int {\left[  {\left( 1 \right)\frac{{\sin 2x}}{2}} \right]} dx \\ \Rightarrow I = \frac{x}{2}\sin 2x -  \frac{1}{2}\int {\sin 2x} dx \\ \end{gathered}


Using the integral rule \int  {\sin kx} dx =  - \frac{{\cos kx}}{k} + c, we have

\begin{gathered} I = \frac{x}{2}\sin 2x - \frac{1}{2}\left(  {\frac{{ - \cos 2x}}{2}} \right) + c \\ \Rightarrow I = \frac{x}{2}\sin 2x +  \frac{1}{4}\cos 2x + c \\ \Rightarrow \int {x\cos 2xdx}  = \frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x + c \\ \end{gathered}

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