# \int x^3 \sqrt{x^2 - 4}dx

## Question:

{eq}\int x^3 \sqrt{x^2 - 4}dx {/eq}

## Integration by substitution :

Some integrals can be solved by using substitution.By substitution we change the variable of integration to a new variable and a simpler function that can be integrated.

Consider {eq}\displaystyle f(x) {/eq}

If the integral is of form {eq}\displaystyle \int \frac{g(x)}{f(x)}dx {/eq}

And {eq}\displaystyle f'(x) =g(x) {/eq}

Let {eq}\displaystyle f(x) =t \Rightarrow f'(x)dx =dt {/eq}

{eq}\displaystyle \int \frac{1}{t}dt {/eq}

Which can be integrated.

Given {eq}\displaystyle I=\int x^3\sqrt{x^2-4}dx {/eq}

Let {eq}\displaystyle x^2-4 =t \Rightarrow 2xdx =dt \Rightarrow xdx = \frac{dt}{2} {/eq}

Substituting we get

{eq}\displaystyle I=\int (t+4)\sqrt{t} dt {/eq}

Rearranging we get

{eq}\displaystyle I= \int t^{\frac{3}{2}}+4t^{\frac{1}{2}} dt {/eq}

Integrating we get

{eq}\displaystyle I= \frac{t^{\frac{5}{2}}}{\frac{5}{2}} +4\frac{t^{\frac{3}{2}}}{\frac{3}{2}} +C {/eq}

That gives

{eq}\displaystyle I=\frac{2}{5}t^{\frac{5}{2}} +\frac{8}{3}t^{\frac{3}{2}}+C {/eq}

So value of integral is given by {eq}\displaystyle I=\frac{2}{5} (x^2-4)^{\frac{5}{2}} +\frac{8}{3}(x^2-4)^{\frac{3}{2}}+C {/eq} 