# Pat made the substitution x - 4 = 6 \sin(t) in an integral and integrated to obtain \int f(x) dx...

## Question:

Pat made the substitution {eq}x - 4 = 6 \sin(t){/eq} in an integral and integrated to obtain {eq}\int f(x) dx = 12 t - 2 \sin(t) \cos(t) + C{/eq}.

Complete Pat's integration by doing the back substitution to find the integral as a function of {eq}x{/eq}

## Integration by substitution

When we make an integration using the substitution method (or variable change), we must at last make also the back substitution. That is equivalent to find the inverse function of the original function used for the variable change. For example, if we use the substitution {eq}y=f(x) {/eq}, then we obtain a solution in function of {eq}y {/eq}. We have to find {eq}f^{-1} {/eq} and make the back substitution {eq}x=f^{-1}(y) {/eq} to express the solution as a function of {eq}x {/eq}

## Answer and Explanation:

For completing Pam's work we must make the inverse function for the back substitution. That is:

{eq}\begin{align} \sin(t)=&\frac{x-4}{6}\\ t&=\arcsin{\frac{x-4}{6}} \end{align} {/eq}

And then we have that:

{eq}\begin{align} \int f(x)dx&=12t-2 \sin(t) \cos(t) + C\\ \int f(x)dx&=12\arcsin(\frac{x-4}{6})-2 \frac{x-4}{6} \sqrt{1-\left(\frac{x-4}{6}\right)^2} + C\\ \end{align} {/eq}

and finally:

{eq}\begin{equation} \int f(x)dx=12 \arcsin\left(\frac{x-4}{6}\right)-\frac{1}{18} (x-4) \sqrt{-x^2+8 x+20} \end{equation} {/eq} 