MHB W.8.7.23 int trig u substitution

[karush]

$\tiny{Whitman \ 8.7.23}$
\begin{align}
\displaystyle
I&=\int \sin^3(t) \cos^2(t) \ d{t} \\
u&=\cos(t) \therefore du=-\sin(t) \, dt \\
\textit{substitute $\cos(t)=u$}&\\
I_u&=-\int (1-u^2) u^2 \, du=-\int(u^2-u^4) \, dt\\
\textit{integrate}&\\
&=-\left[\frac{u^3}{3}-\frac{u^5}{5}\right]
=\left[\frac{u^5}{5}-\frac{u^3}{3}\right]\\
\textit{back substitute $u=\cos(t)$}&\\
I&= \frac{\cos^5(t)}{5}-\frac{\cos^3(t)}{3}+C
\end{align}
$\textit{think this is correct but suggestions??}$
$\textit{also how is text like "substitute" left justifed withinn an align?}$

 

[Greg]

That's correct and probably the most efficient method. As for the LaTeX, double-slashes generate a new line.

 

[MarkFL]

I would use the \begin \end environment instead:

$$\begin{array}{lr} & u=\cos(t)\therefore du=-\sin(t) \\ \text{substitute }\cos(t)=u & \end{array}$$

 

[karush]

That's correct and probably the most efficient method. As for the LaTeX, double-slashes generate a new line.

thanks..
I have a really hard time with u subsubsection ,, especially choosing one that will require the least amount of steps. the best examples have always been here...