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…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter2: Equations And Inequalities

Section2.5: Other Types Of Equations

Problem 55E

See similar textbooks

Related questions

Concept explainers

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

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

Step by step

Solved in 3 steps

Knowledge Booster

$Definite Integral$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

Similar questions