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}