Question

In Exercises \(25-46,\) use substitution to evaluate the integral. $$\int \frac{d x}{\sin ^{2} 3 x}$$

Short answer

-\frac{1}{3} \cot 3x + c

Step-by-step solution

Choose Appropriate Substitution

In our case, let's choose \(u = 3x\). This simplifies the integral and its subsequent steps.

Calculate Differential of u

Differentiating \(u = 3x\) we get, \(du = 3dx\). To isolate \(dx\) we then divide both sides by 3, \(dx = \frac{du}{3}\).

Substitute u and dx in the Integral

Now, substitute \(u\) and \(dx\) into the integral. The integral \(\int \frac{dx}{\sin^{2}3x}\) becomes \(\int \frac{1}{\sin^{2}u} * \frac{du}{3}\). We factor out constants from the integral, giving us \(\frac{1}{3} \int \frac{du}{\sin^{2}u}\)

Evaluate the Integral

The integral \(\frac{1}{3} \int \frac{du}{\sin^{2}u}\) is a standard integral which equates to \(-\frac{1}{3} \cot u + c\), where 'c' is the constant of integration.

Back Substitute u Into the Answer

Substituting \(u = 3x\) into \(-\frac{1}{3} \cot u + c\), we get our final answer as \(-\frac{1}{3} \cot 3x + c\).