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Chapter 5: Q 41. (page 452)

Solve the following integral.
$\int \sin^{2} 3 x \cos^{2} 3 x d x$

Short Answer

Expert verified

Answer is $\frac{x}{8} - \frac{\sin 12 x}{96} + C$

Step by step solution

01

Step 1. Given information

An integral is $\int \sin^{2} 3 x \cos^{2} 3 x d x$

02

Step 2. Explanation

Let $u = 3 x$

$\begin{array}{rcl} \int \sin^{2} 3 x \cos^{2} 3 x d x & = & \frac{1}{3} \int \sin^{2} u \cos^{2} u d u \\ = & \frac{1}{12} \int \sin^{2} 2 u d u \\ = & \frac{1}{24} \int 1 - \cos 4 u d u \\ = & \frac{1}{24} (u - \frac{\sin 4 u}{4}) + C \\ = & \frac{1}{24} (3 x - \frac{\sin 12 x}{4}) + C \\ = & \frac{x}{8} - \frac{\sin 12 x}{96} + C \end{array}$

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Most popular questions from this chapter

Solve given definite integral.

$\int_{1 / 4}^{1 / 2} \frac{1}{x^{2} \sqrt{1 - x^{2}}} d x$

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

For each function u(x) in Exercises 9–12, write the differential du in terms of the differential dx.

$u (x) = x^{2} + x + 1$

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.

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

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

$2 x^{2} - 4 x + 1$

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