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Chapter 5: Problem 50

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{\pi / 16}^{\pi / 8} 8 \csc ^{2} 4 x d x$$

Short Answer

Expert verified
Question: Evaluate the definite integral of the function $$8 \csc ^{2} 4 x$$ from $$\pi / 16$$ to $$\pi / 8$$. Answer: $$\int_{\pi / 16}^{\pi / 8} 8 \csc^2 4 x d x = 8$$.

Step by step solution

01

Find the antiderivative

To find the antiderivative of $$8\csc^2{4x}$$, we should recall that the derivative of $$-\cot{x}$$ is $$\csc^2{x}$$. Therefore, the antiderivative of $$8\csc^2{4x}$$ is $$-8\cot(4x)$$, taking into account the chain rule.
02

Use the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, we find the definite integral by evaluating the antiderivative at the given limits and taking the difference. $$\int_{\pi / 16}^{\pi / 8} 8 \csc^2 4 x dx = -8\cot(4x)\Big|_{\pi/16}^{\pi/8}$$
03

Substitute the limits of integration

Substitute the limits of integration and find the difference. $$-8\cot(4x)\Big|_{\pi/16}^{\pi/8} = -8\left(\cot\left(\frac{4\pi}{8}\right) - \cot\left(\frac{4\pi}{16}\right)\right)$$ $$= -8\left(\cot\left(\frac{\pi}{2}\right) - \cot\left(\frac{\pi}{4}\right)\right)$$
04

Simplify the expression

Using the fact that $$\cot(\pi/2) = 0$$ and $$\cot(\pi/4) = 1$$, we find the final answer. $$= -8 \left(0 - 1\right) = \boxed{8}$$ So, $$\int_{\pi / 16}^{\pi / 8} 8 \csc^2 4 x d x = 8$$.

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