Question Statement

Evaluate the integral: $\int (\cot 9y-3)dy$

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Background and Explanation

This problem requires integrating a trigonometric expression using substitution techniques. You should be familiar with the quotient identity for cotangent and the logarithmic integration rule for derivatives over functions.

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Solution

We begin by applying the linearity property of integration to split the integral into two separate terms:

$\begin{array}{rl}I& =\int (\cot 9y-3)dy\\ & =\int \cot 9ydy-3\int 1dy\end{array}$

Next, we rewrite $\cot 9y$ using the identity $\cot \theta =\frac{\cos \theta }{\sin \theta }$:

$I=\int \frac{\cos 9y}{\sin 9y}dy-3y$

To integrate the first term, we recognize that the numerator should contain the derivative of the denominator ($\frac{d}{dy}(\sin 9y)=9\cos 9y$). We multiply and divide by 9 to create this form:

$I=\frac{1}{9}\int \frac{9\cos 9y}{\sin 9y}dy-3y$

Now we apply the logarithmic integration formula $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln (f(x))$, where $f(y)=\sin 9y$:

$I=\frac{1}{9}\ln (\sin 9y)-3y+c$

Verification

To confirm our answer, we differentiate the result with respect to $y$:

$\begin{array}{rl}\frac{d}{dy}\left(\frac{1}{9}\ln (\sin 9y)-3y+c\right)& =\frac{1}{9}\cdot \frac{1}{\sin 9y}\cdot \frac{d}{dy}(\sin 9y)-3\frac{d}{dy}(y)+\frac{d}{dy}(c)\\ & =\frac{1}{9}\cdot \frac{1}{\sin 9y}\cdot 9\cos 9y-3\\ & =\cot 9y-3\end{array}$

Since differentiation returns the original integrand, the solution is verified.

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Key Formulas or Methods Used

• Linearity of Integration: $\int (f(y)-g(y))dy=\int f(y)dy-\int g(y)dy$
• Trigonometric Identity: $\cot \theta =\frac{\cos \theta }{\sin \theta }$
• Logarithmic Integration Rule: $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln (f(x))+C$
• Chain Rule: $\frac{d}{dy}[\ln (u)]=\frac{1}{u}\cdot \frac{du}{dy}$

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Summary of Steps

1. Separate the integral into $\int \cot 9ydy$ and $-3\int dy$ using linearity
2. Rewrite $\cot 9y$ as $\frac{\cos 9y}{\sin 9y}$ using the quotient identity
3. Multiply and divide by 9 to match the derivative pattern $\frac{f^{\prime }(y)}{f(y)}$
4. Apply logarithmic integration to obtain $\frac{1}{9}\ln (\sin 9y)$
5. Integrate the constant term to get $-3y$ and add the constant of integration $+c$
6. Verify by differentiation using the chain rule to confirm the result matches the original integrand