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Chapter 5: Q 4. (page 495)

Use Pythagorean identities to rewrite each of the following trigonometric expressions.
$\sec^{6} x =_____$

Short Answer

Expert verified

The obtained result is $\frac{1}{{(1 - \sin^{2} x)}^{3}}$ .

Step by step solution

01

Step 1. Given information.

Consider the given expression.

$\sec^{6} x$

02

Step 2. Apply the Pythagorean identities to rewrite the expression.

The Pythagorean identity is defined as,

$\sin^{2} x + \cos^{2} x = 1$

Use the identity to write the given expression.

$\begin{array}{rcl} \cos^{2} x & = & 1 - \sin^{2} x \\ {(\cos^{2} x)}^{3} & = & {(1 - \sin^{2} x)}^{3} \\ \cos^{6} x & = & {(1 - \sin^{2} x)}^{3} \\ \frac{1}{\cos^{6} x} & = & \frac{1}{{(1 - \sin^{2} x)}^{3}} \\ \sec^{6} x & = & \frac{1}{{(1 - \sin^{2} x)}^{3}} \end{array}$

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Most popular questions from this chapter

Solve given integrals by using polynomial long division to rewrite the integrand. This is one way that you can sometimes avoid using trigonometric substitution; moreover, sometimes it works when trigonometric substitution does not apply.

$鈭� \frac{x^{4} - 3}{2 + 3 x^{2}} d x$

Show by differentiating (and then using algebra) that $鈭� \cot 鈦� (\sin^{鈭� 1} 鈦� x)$ and $鈭� \frac{\sqrt{1 鈭� x^{2}}}{x}$ are both antiderivatives of $\frac{1}{x^{2} \sqrt{1 鈭� x^{2}}}$ . How can these two very different-looking functions be an antiderivative of the same function?

Explain why, if $x = a s e c u$ , then $\sqrt{a^{2} s e c^{2} u 鈭� a^{2}}$ is $鈭� a \tan u$ if $x < 鈭� a$ and is $a \tan u$ if $x > a$ . Your explanation should include a discussion of domains and absolute values.

Solve given definite integral.

${鈭�}_{4}^{5} 鈥� \frac{1}{x \sqrt{x^{2} + 9}} d x$

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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