MHB Is the Tabular Method the Easiest Way to Solve Integrals?

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The discussion centers around evaluating the integral I = ∫ x csc²(6x) dx, with participants sharing their methods. One user presents a solution using integration by parts, correctly identifying u and dv, leading to the final result involving a logarithmic term. Another user mentions the tabular method as a potentially easier approach but admits to not fully understanding it. The tabular method is described as organizing u and its derivatives in one column and dv and its antiderivatives in another, which some find more useful. Overall, the conversation highlights different techniques for solving integrals, with interest in the efficiency of the tabular method.
karush
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$\textsf{1) Evaluate the integral}$
\begin{align*}
I_1&=\int x \csc^2 6x \, dx\\
u&=6x\therefore du=dx\\
dv&=\csc^2(6x) \, dx \therefore v=-\frac{1}{6}\cot(6x) \\
u&=6x\therefore du=6 \, dx \\
I&=-\frac{1}{6} x\cot(6x)+\frac{1}{36}\int \cot(u) \, dx\\
&=\color{red}{
-\frac{1}{6} x\cot(6x)
+\frac{1}{36}\ln(|\sin36)|)}
\end{align*}

Ok I think this is correct, if so
saw another student use the tabular method to solve this
but couldn't see good enought to understand it
is that a lot easier.:cool:
 
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I think you've gotten your $u$'s mixed up. We are given:

$$I=\int x\csc^2(6x)\,dx$$

Let:

$$u=x\implies du=dx$$

$$dv=\csc^2(6x)\,dx\implies v=-\frac{1}{6}\cot(6x)$$

Hence:

$$I=-\frac{x}{6}\cot(6x)+\frac{1}{6}\int \cot(6x)\,dx=\frac{1}{36}\ln\left|\sin(6x)\right|-\frac{x}{6}\cot(6x)+C$$

I don't know anything about the "tabular method." :D
 
tabular integration ...

$u$ and its derivatives in the second column

$dv$ and its antiderivatives in the third column

View attachment 6450
 
Last edited by a moderator:
well that is certainly much more usefull

I guess the uv method was to illustrate how it works short of a full proof

like your chart!
 
This method is also called the "Stand and Deliver" method, because it was featured in that movie.
 

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