Question

Solve given definite integral.

${\int }_{-1}^1 x^3\sqrt{9x^2-1}dx$

Short answer

$0$

Step-by-step solution

Step1. Given Information

The integral is as follows.

${\int }_{-1}^1 x^3\sqrt{9x^2-1}dx$

The objective is to solve the integral.

Step2. Assumptions

To solve the integral, let $u=x^2$

Now, derivation of above equation is solved below.

$$\begin{gathered} u=x^2 \\ du=2xdx \end{gathered}$$

Step3. Solution

The integral is solved below.

$\begin{array}{r}\int x^3\sqrt{9x^2-1}dx\\ =\int \frac{u}{2}\sqrt{9u-1}du\\ =\frac{1}{2}\int u\sqrt{9u-1}du\end{array}$

$=\frac{1}{2}\int \frac{2}{9}{v}^2\left(\frac{v^2}{9}+\frac{1}{9}\right)dv\left[v=\sqrt{9u-1},dv=\frac{9}{2\sqrt{9u-1}}dx\right]$

$=\frac{1}{9}\int \frac{v^2}{9}\left(v^2+1\right)dv$

$=\frac{1}{81}\int \left(v^4+v^2\right)dv$

$=\frac{1}{81}\int v^{4}dv+\frac{1}{81}\int v^{2}dv$

$=\frac{1}{405}(v{)}^5+\frac{1}{243}(v{)}^3$

$=\frac{1}{405}(\sqrt{9u-1}{)}^5+\frac{1}{243}(\sqrt{9u-1}{)}^3$

$=\frac{1}{405}{\left(\sqrt{9x^2-1}\right)}^5+\frac{1}{243}{\left(\sqrt{9x^2-1}\right)}^3$

$=\left(\frac{x^2}{45}+\frac{2}{1215}\right){\left(9x^2-1\right)}^{\frac{3}{2}}$

Step4. Solution

The definite integral is solved below.

${\int }_{-1}^1 x^3\sqrt{9x^2-1}dx$

$={\left[\left(\frac{x^2}{45}+\frac{2}{1215}\right){\left(9x^2-1\right)}^{\frac{3}{2}}\right]}_{-1}^1$

$=\left[\left(\frac{1^2}{45}+\frac{2}{1215}\right){\left(9(1{)}^2-1\right)}^{\frac{3}{2}}\right]-\left[\left(\frac{(-1{)}^2}{45}+\frac{2}{1215}\right){\left(9(-1{)}^2-1\right)}^{\frac{3}{2}}\right]$

$=0$

Therefore, the value is 0