Evaluate the integral by making the given substitution.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^2\sqrt[\phantom{2}]{x^{3}36}dx,y=x^{3}36$

Step $1$

We know that if $u=f(x),$ then $du=f^{\text{'}}(x)dx.$ Therefore, if $u=x^{3}36,$ then $dv=$

$3x^2$

$dx$

Step $2$

If $u=x^{3}36$ is substituted into ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^2\sqrt[\phantom{2}]{x^{3}36}dx,$ then we have ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^2(u{)}^{\frac{1}{2}}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{u}^{\frac{1}{2}}{x}^{2}dx.$

We must also convert $x^2$ dx into an expression involving $u.$

We know that $du=3x^{2}dx,$ so $x^{2}dx=$

$\frac{1}{3}$ da

Dieg $3$

Now, $xv=x^{3}36,$ then ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^2\sqrt[\phantom{2}]{x^{3}36}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{v}^{\frac{1}{2}}(\frac{1}{3}dv)=\frac{1}{3}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{v}^{\frac{1}{2}}dv.$

This evaluates as follows. $($Enter your answer in terma of u$.)$

$\frac{1}{2}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{y}^{\frac{1}{2}}dx=c$