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Chapter 7: Problem 38

Evaluate the following integrals. $$\int \sec ^{-2} x \tan ^{3} x d x$$

Short Answer

Expert verified
Question: Evaluate the integral $\int \sec ^{-2} x \tan ^{3} x d x$. Answer: $\int \sec ^{-2} x \tan ^{3} x d x = -\sec x + C$

Step by step solution

01

Rewrite the integral in terms of sine and cosine functions

The given integral can be rewritten as follows: $$\int \sec ^{-2} x \tan ^{3} x d x = \int \frac{\sin^3x}{\cos^5x}dx$$
02

Perform a substitution

Let's perform a substitution to simplify the integral. Let \(u = \cos x\). Then, \(-du = \sin x dx\). The integral becomes: $$\int \frac{-u^3}{u^5}(-du) = \int \frac{u^3}{u^5} du$$
03

Simplify the integral

Now, we simplify the integral: $$\int \frac{u^3}{u^5} du = \int u^{-2} du$$
04

Evaluate the integral

Now, we can integrate with respect to \(u\): $$\int u^{-2} du = \frac{u^{-1}}{-1} + C = -u^{-1} + C$$
05

Undo the substitution

Now, let's undo the substitution and obtain our final answer. Recall that \(u = \cos x\). Therefore, our final answer is: $$-u^{-1} + C = -\frac{1}{\cos x} + C = -\sec x + C$$ Our final solution is: $$\int \sec ^{-2} x \tan ^{3} x d x = -\sec x + C.$$

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