MHB W.8.7.23 int trig u substitution

karush
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$\tiny{Whitman \ 8.7.23}$
\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}
$\textit{think this is correct but suggestions??}$
$\textit{also how is text like "substitute" left justifed withinn an align?}$
 
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That's correct and probably the most efficient method. As for the LaTeX, double-slashes generate a new line.
 
I would use the \begin \end environment instead:

$$\begin{array}{lr} & u=\cos(t)\therefore du=-\sin(t) \\ \text{substitute }\cos(t)=u & \end{array}$$
 
greg1313 said:
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...😎
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...

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