Question

Find the area of the region. Use a graphing utility to verify your result. $$ \int_{\pi / 12}^{\pi / 4} \csc 2 x \cot 2 x d x $$

Short answer

The solution to the integral is the difference between the evaluated antiderivative at \(\pi / 4\) and at \(\pi / 12 \). It represents the area of the region between the curve and the x-axis from \(\pi / 12 \) to \(\pi / 4\) under the curve f(x) = \(\csc 2 x \cot 2 x \). As this exercise also suggests using a graphing utility to verify the result, plot the function \( \csc 2 x \cot 2 x \) and identify the area under the curve from \(\pi / 12 \) to \( \pi / 4 \) to confirm the calculated result.

Step-by-step solution

Simplify the Integrand

Note that cot(x) can be written as csc^2(x) - 1. Apply this identity and the integrand becomes: \(\csc(2x) * (\csc^2(2x) - 1)\). Multiply and we obtain: \(\csc^3(2x) - \csc(2x)\).

Integrate

The antiderivative of csc(x) is -ln|csc(x) + cot(x)|, and we can use this to calculate the antiderivative of our integrand. Integrating each term separately, we obtain: -1/2 ln|csc(2x) + cot(2x)|^3 - ln|csc(2x) + cot(2x)| = -1/2 ln|csc^2(2x) + 2csc(2x)cot(2x) + cot^2(2x)| - ln|csc(2x) + cot(2x)|.

Apply the Limits of Integration

Now, apply the limits of integration \(\pi / 12\) to \(\pi / 4\). Calculate the value of the antiderivative at these points, subtracting the value at the lower limit from the value at the upper: \[ \left[ -\frac{1}{2} \ln |\csc^2(2x) + 2csc(2x)cot(2x) + cot^2(2x)| - \ln |\csc(2x) + cot(2x)| \right]_{\frac{\pi}{12}}^{\frac{\pi}{4}} \]

Evaluate the Definite Integral

Substitute \( x = \pi / 4 \) and \(x = \pi / 12\) into the definite integral. Subtract the value at \(\pi / 4\) from the value at \(\pi / 12\) (upper limit minus lower limit). The result is the area of the region between the curve and the x-axis from \(\pi / 12\)to \(\pi / 4\) under the curve of \(\csc 2 x \cot 2 x \).