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Chapter 7: Problem 39

Evaluate the following integrals. $$\int \frac{\csc ^{4} x}{\cot ^{2} x} d x$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int \frac{\csc ^{4} x}{\cot ^{2} x} d x$$ Answer: The integral evaluates to $$\frac{1}{\cos x} + 2\cos x - \frac{\cos^{3} x}{3} + C$$ where C is the constant of integration.

Step by step solution

01

Rewrite the integral using trigonometric identities

We are given the integral: $$\int \frac{\csc ^{4} x}{\cot ^{2} x} d x$$ Recall that \(\csc x = \frac{1}{\sin x}\) and \(\cot x = \frac{\cos x}{\sin x}\). Hence, we can rewrite our given integral as: $$\int \frac{\left(\frac{1}{\sin x}\right)^{4}}{\left(\frac{\cos x}{\sin x}\right)^{2}} d x$$ Now, simplifying the integral further: $$\int \frac{\sin^{4}x}{\cos^{2}x} d x$$
02

Apply substitution

Let's make a substitution, \(u = \cos x\). Then, note that \(-du = \sin x dx\). Now the integral becomes: $$\int \frac{(1-u^{2})^{2}}{u^{2}}(-du) $$ Simplifying the integral: $$-\int \frac{(1-2u^{2}+u^{4})}{u^{2}} du $$
03

Break the integral into parts

Next, let's break the integral into the sum of simpler integrals: $$-\left(\int \frac{1}{u^{2}} du - 2\int u^{2-2} du + \int u^{4-2} du \right)$$ Now we can integrate each part of the integral: $$-\left(\int u^{-2} du - 2\int u^{0} du + \int u^{2} du \right)$$
04

Perform integration

Evaluating each part, we get: $$-\left(-\frac{1}{u} - 2u + \frac{u^{3}}{3} \right) + C$$ Where C is the constant of integration. Now, let's substitute back \(u = \cos x\). We get: $$-\left(-\frac{1}{\cos x} - 2\cos x + \frac{\cos^{3} x}{3} \right) + C$$ Our final answer is: $$\frac{1}{\cos x} + 2\cos x - \frac{\cos^{3} x}{3} + C$$

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