# Using the reduction formula, find the integral of integral x^n cos (x) dx in terms of an integral...

## Question:

Using the reduction formula, find the integral of {eq}\displaystyle \int x^n \cos (x)\ dx {/eq} in terms of an integral involving {eq}x^{n - 1} {/eq}.

## Integration by Reduction Formula.

Reduction formula reduces an integral from higher to lower power with the same form.

Also reduction formula repeated continuously until it forms an easier integration.

The formula is:

{eq}\displaystyle\int cosx\ dx=sinx+c\\\\ \displaystyle\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+c\\\\ \displaystyle\int sinx\ dx=-cosx+c\\\\ {/eq}

We have to integrate the given equation by using reduction formula:

{eq}\displaystyle\int x^ncos(x)dx\\\\ I_n=\displaystyle\int x^ncos(x)dx\\\\ {/eq}

Applyin formula of integration by parts:

{eq}\displaystyle\int (uv)dx=u\displaystyle\int v\ dx-\displaystyle\int \left [ \dfrac{d}{dx}(u)\displaystyle\int v\ dx \right ]dx\\\\ =x^n\displaystyle\int cosx\ dx-\displaystyle\int \left [ \dfrac{d}{dx}(x^n)\displaystyle\int cosx\ dx \right ]dx\\\\ =x^n\ sinx-\displaystyle\int \left [ nx^{n-1}\ sinx \right ]dx\\\\ =x^n\ sinx-n\displaystyle\int x^{n-1}sinx\ dx\\\\ {/eq}

Here the equation is going towards non ending loop.

Again integrating:

{eq}=x^n\ sinx-n\left [ x^{n-1}\displaystyle\int sinx\ dx-\displaystyle\int \left [ \dfrac{d}{dx}(x^{n-1})\displaystyle\int sinx\ dx \right ] \right ]dx\\\\ =x^n\ sinx-n\left [ x^{n-1}(-cosx)-\displaystyle\int \left [ (n-1)x^{n-2}(-cosx) \right ]dx \right ]\\\\ =x^n\ sinx+nx^{n-1}cosx-n(n-1)\displaystyle\int x^{n-2}cosx\ dx\\\\ =x^n\ sinx+nx^{n-1}cosx-n(n-1)\ I_{n-2}\\\\ {/eq}

This is the required reduction formula.