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Chapter 5: Q. 73 (page 452)

Solve each of the definite integrals in Exercises 67鈥�76.
${鈭�}_{0}^{\frac{蟺}{4}} s e c^{4} x \tan^{2} x d x$ .

Short Answer

Expert verified

The answer is $\frac{8}{15}$ .

Step by step solution

01

Step 1. Given information.

The integral is ${鈭�}_{0}^{\frac{蟺}{4}} s e c^{4} x \tan^{2} x d x$ .

02

Step 2. Explanation.

Rewrite the integral and substitute $\tan^{2} x + 1 = s e c^{2} x$ .

${鈭�}_{0}^{\frac{蟺}{4}} s e c^{2} x s e c^{2} x \tan^{2} x d x = {鈭�}_{0}^{\frac{蟺}{4}} (1 + \tan^{2} x) s e c^{2} x \tan^{2} x d x$

Substitute $u = t a n x$ . Differentiate it.

$d u = s e c^{2} x d x$

limit changes to 0 to 1.

${鈭�}_{0}^{1} (1 + u^{2}) u^{2} d u = {鈭�}_{0}^{1} u^{2} + u^{4} d u = {[\frac{u^{3}}{3}]}_{0}^{1} + {[\frac{u^{5}}{5}]}_{0}^{1} = \frac{{(1)}^{3} - 0^{3}}{3} + \frac{1^{5} - 0^{5}}{5} = \frac{1}{3} + \frac{1}{5} = \frac{8}{15}$

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Most popular questions from this chapter

Explain why it makes sense to try the trigonometric substitution $x = \sec 鈦� u$ if an integrand involves the expression $\sqrt{x^{2} 鈭� 1}$

Solve the integral: $鈭� \frac{x}{e^{x}} d x$

Which of the integrals that follow would be good candidates for trigonometric substitution? If a trigonometric substitution is a good strategy, name the substitution. If another method is a better strategy, explain that method.

$(a) 鈭� \frac{4 + x^{2}}{x} d x$ $(b) 鈭� \frac{x}{4 + x^{2}} d x$

role="math" localid="1648759296940" $(c) 鈭� \frac{x^{2}}{4 + x^{2}} d x$ $(d) 鈭� \frac{16 鈭� x^{4}}{4 + x^{2}} d x$

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.

$2 {(x + 2)}^{2}$

Solve the integral: $鈭� \ln \sqrt{x} d x$

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