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Chapter 5: Q 29. (page 452)

Solve the following integral.
$鈭� c o t^{5} x d x$

Short Answer

Expert verified

Answer is $- \frac{c o t^{4} x}{4} + \frac{c o t^{2} x}{2} - \frac{\ln (c o t^{2} x + 1)}{2} + C$

Step by step solution

01

Step 1. Given information

An integral is $鈭� c o t^{5} x d x$

02

Step 2. Explanation

Let $c o t x = u$

\begin{array}{rcl} 鈭� c o t^{5} x d x & = & 鈭� \frac{- u^{5}}{u^{2} + 1} d u \\ = & - 鈭� u^{3} - u + \frac{u}{u^{2} + 1} d u \\ = & - \frac{u^{4}}{4} + \frac{u^{2}}{2} - \frac{\ln (u^{2} + 1)}{2} + C \\ = & - \frac{c o t^{4} x}{4} + \frac{c o t^{2} x}{2} - \frac{\ln (c o t^{2} x + 1)}{2} + C \end{array}

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Most popular questions from this chapter

Explain why, if $x = a \sin u$ , then $\sqrt{a^{2} 鈭� x^{2}} = a \cos u$ . Your explanation should include a discussion of domains and absolute values.

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$鈭� \sqrt{3 - x} \sqrt{x - 1} d x$

Solve the integral

$鈭� \sqrt{4 - x} \sqrt{x - 2} d x$

Complete the square for each quadratic in Exercises 28鈥�33. Then describe the trigonometric substitution that would be appropriate if you were solving an integral that involved that quadratic.

$x^{2} + 6 x 鈭� 2$

Suppose $u (x) = x^{2}$ . Calculate and compare the values of the following definite integrals:

role="math" localid="1648786835678" ${鈭�}_{- 1}^{5} u^{2} d u, {鈭�}_{x = - 1}^{x = 5} u^{2} d u and {鈭�}_{u (- 1)}^{u (5)} u^{2} d u$

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