Evaluate the integral. integral sin 6 x dx

Given ∫sin6x dx We use sin2x=1-cos2x2cos2x=1+cos2x2

Answered: Evaluate the integral. integral sin 6…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

Evaluate the integral.

integral sin ⁶ x dx

Expert Solution

Step 1

Given

$\int \sin^{6} x d x$

We use

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

Step 2

We can write

$\begin{array}{rcl} \sin^{6} x & = & {(\sin^{2} x)}^{2} \sin^{2} x \\ = & {(\frac{1 - \cos 2 x}{2})}^{2} (\frac{1 - \cos 2 x}{2}) \\ = & \frac{1}{8} (1 - 2 \cos 2 x + \cos^{2} 2 x) (1 - \cos 2 x) \\ = & \frac{1}{8} (1 - 2 \cos 2 x + \frac{1 + \cos 4 x}{2}) (1 - \cos 2 x) \\ = & \frac{1}{16} (2 - 4 \cos 2 x + 1 + \cos 4 x) (1 - \cos 2 x) \\ = & \frac{1}{16} (3 - 4 \cos 2 x + \cos 4 x) (1 - \cos 2 x) \\ = & \frac{1}{16} (3 - 4 \cos 2 x + \cos 4 x - 3 \cos 2 x + 4 \cos^{2} 2 x - \cos 2 x \cos 4 x) \\ = & \frac{1}{16} (3 - 7 \cos 2 x + \cos 4 x + 2 (1 + \cos 4 x) - \frac{1}{2} (\cos 6 x + \cos 2 x)) \\ = & \frac{1}{16} (3 - 7 \cos 2 x + \cos 4 x + 2 + 2 \cos 4 x - \frac{1}{2} (\cos 6 x + \cos 2 x)) \\ = & \frac{1}{16} (3 - 7 \cos 2 x + 3 \cos 4 x + 2 - \frac{1}{2} (\cos 6 x + \cos 2 x)) \\ = & \frac{1}{32} (6 - 14 \cos 2 x + 6 \cos 4 x + 4 - \cos 6 x - \cos 2 x) \\ = & \frac{1}{32} (10 - 15 \cos 2 x + 6 \cos 4 x - \cos 6 x) \end{array}$

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