# Find the following anti-derivative. integral x^2 \cos(x^3 - 5) dx

## Question:

Find the following anti - derivative.

{eq}\displaystyle \int x^2 \cos(x^3 - 5) dx {/eq}

## Integration by substitution:

The method of substitution is used when it is complicate to integrate the given integrand directly.

If the given integral is in the form or can be made in the form,

{eq}\displaystyle \int s(t(x)) \cdot t'(x) \ dx {/eq},

can be written as

{eq}\displaystyle \int s(u) \ du {/eq},

where {eq}u = t(x) {/eq} and {eq}du = t'(x) dx {/eq}.

We have,

{eq}\displaystyle \int x^2 \cos(x^3 - 5) dx {/eq}

Rewrite as,

{eq}\displaystyle \dfrac {3}{3} \int x^2 \cos(x^3 - 5) dx = \dfrac {1}{3} \int \cos(x^3 - 5) \cdot 3x^2 dx {/eq}

Now use the method of substitution,

Let {eq}u = x^3 - 5 \Rightarrow du = 3x^2 \ dx {/eq}

Then,

{eq}\displaystyle \begin{align*} \int x^2 \cos(x^3 - 5) dx &= \dfrac {1}{3} \int \cos u \ du \\ &= \dfrac {1}{3} \cdot \sin u \\ &= \dfrac {\sin u }{3} \\ \end{align*} {/eq}

Substitute back {eq}u = x^3 - 5 {/eq} and add the constant of integration,

{eq}\Rightarrow \displaystyle \int x^2 \cos(x^3 - 5) dx = \dfrac {\sin (x^3 - 5)}{3} + C {/eq}