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Chapter 5: Q. 67 (page 452)

Solve each of the definite integrals in Exercises 67鈥�76.
${鈭�}_{0}^{蟺} \sin^{5} (x) d x$

Short Answer

Expert verified

The solution of the integral is $\frac{16}{15}$ .

Step by step solution

01

Step 1. Given Information

The given integral is ${鈭�}_{0}^{蟺} \sin^{5} (x) d x$ .

02

Step 2. Rewrite and substitute

USe the trigonometric identities to rewrite the given integral as follows:

${鈭�}_{0}^{蟺} \sin^{5} (x) d x = {鈭�}_{0}^{蟺} {(1 - \cos^{2} (x))}^{2} \sin (x) d x$

Assume that $\cos (x) = 0$ . So, $- \sin (x) d x = d u$ .
Substitute inte the integral and then simplify.

$\begin{array}{rcl} {鈭�}_{0}^{蟺} {(1 - \cos^{2} (x))}^{2} \sin (x) d x & = & - {鈭�}_{\cos^{- 1} 0}^{\cos^{- 1} 蟺} {(1 - u^{2})}^{2} d u \\ = & {鈭�}_{1}^{- 1} - 1 + 2 u^{2} - u^{4} d u \\ = & {鈭�}_{- 1}^{1} 1 - 2 u^{2} + u^{4} d u \end{array}$

03

Step 3. Integrate

Integrate the obtained integral and then substitute the limits.

$\begin{array}{rcl} {鈭�}_{- 1}^{1} 1 - 2 u^{2} + u^{4} d u & = & {[u - \frac{2}{3} u^{3} + \frac{u^{5}}{5}]}_{- 1}^{1} \\ = & 1 - \frac{2}{3} + \frac{1}{5} + 1 - \frac{2}{3} + \frac{1}{5} \\ = & \frac{16}{15} \end{array}$

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Most popular questions from this chapter

For each integral in Exercises 5鈥�8, write down three integrals that will have that form after a substitution of variables.

$鈭� u^{2} d u$

Solve the integral: $鈭� x \ln (x^{2}) d x$

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" $(c) 鈭� \frac{x^{2}}{4 + x^{2}} d x$ $(d) 鈭� \frac{16 鈭� x^{4}}{4 + x^{2}} d x$

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

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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