Question Statement

Evaluate the indefinite integral:

$\int {\csc }^2\left(\frac{x-1}{3}\right)dx$

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Background and Explanation

This problem requires evaluating an indefinite integral involving a composite trigonometric function where the argument is a linear expression. The key prerequisites are the method of substitution (to handle the inner function $\frac{x-1}{3}$) and the standard integral formula for ${\csc }^{2}x$.

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Solution

Let $I$ denote the integral:

$I=\int {\csc }^2\left(\frac{x-1}{3}\right)dx$

To simplify the composite function, we use substitution. The inner linear function $\frac{x-1}{3}$ suggests setting:

$t=\frac{x-1}{3}$

Now we find the relationship between $dx$ and $dt$ by differentiating. Multiplying both sides by 3:

$x-1=3t$

Solving for $x$:

$x=3t+1$

Differentiating with respect to $t$:

$\begin{array}{rl}\frac{dx}{dt}& =3\\ \Rightarrow dx& =3dt\end{array}$

Substituting $t=\frac{x-1}{3}$ and $dx=3dt$ back into the original integral:

$I=\int {\csc }^{2}t\cdot 3dt$

Factor out the constant 3:

$I=3\int {\csc }^{2}tdt$

Using the standard integral formula $\int {\csc }^{2}xdx=-\cot x+c$:

$\begin{array}{rl}I& =3(-\cot t)+c\left(\because \int {\csc }^{2}xdx=-\cot x\right)\\ & =-3\cot t+c\end{array}$

Finally, substitute back $t=\frac{x-1}{3}$ to express the answer in terms of $x$:

$I=-3\cot \left(\frac{x-1}{3}\right)+c$

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Key Formulas or Methods Used

• Integration by Substitution: $\int f(g(x))\cdot g^{\prime }(x)dx=\int f(u)du$ where $u=g(x)$
• Standard Trigonometric Integral: $\int {\csc }^{2}xdx=-\cot x+c$ (equivalently $\int {\mathrm{cosec}}^{2}xdx=-\cot x+c$)
• Derivative of Linear Function: $\frac{d}{dx}\left(\frac{x-1}{3}\right)=\frac{1}{3}$, which gives $dx=3dt$ when substituting $t=\frac{x-1}{3}$

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Summary of Steps

1. Substitute: Let $t=\frac{x-1}{3}$ to simplify the argument of the cosecant function
2. Differentiate: Compute $dx=3dt$ from the substitution equation
3. Transform: Rewrite the integral as $I=3\int {\csc }^{2}tdt$
4. Integrate: Apply the standard formula to obtain $-3\cot t+c$
5. Back-substitute: Replace $t$ with $\frac{x-1}{3}$ to get the final result $-3\cot \left(\frac{x-1}{3}\right)+c$