Use the half-angle formula cos² 0 = 16 V cos² cos2 Ꮎ ᏧᎾ 6 = 1/²/(1+ = 1/6/6/²2 ( ¹ + [ 1 8 ([ 1) + C to evaluate the integral in terms of 0. ) de
Use the half-angle formula cos² 0 = 16 V cos² cos2 Ꮎ ᏧᎾ 6 = 1/²/(1+ = 1/6/6/²2 ( ¹ + [ 1 8 ([ 1) + C to evaluate the integral in terms of 0. ) de
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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![Use the half-angle formula \(\cos^2 \theta = \frac{1}{2} \left(1 + \boxed{\phantom{u}}\right)\) to evaluate the integral in terms of \(\theta\).
\[
\frac{16}{\sqrt{6}} \int \cos^2 \theta \, d\theta = \frac{16}{\sqrt{6}} \int \frac{1}{2} \left(1 + \boxed{\phantom{u}}\right) \, d\theta
\]
\[
= \frac{8}{\sqrt{6}} \left( \boxed{\phantom{u}} \right) + C
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2065d956-6c8d-4dd3-a865-df08e0123eae%2Fb6444d7f-711c-428b-a259-5b68fd4dc38e%2Ffqgoi7i_processed.png&w=3840&q=75)
Transcribed Image Text:Use the half-angle formula \(\cos^2 \theta = \frac{1}{2} \left(1 + \boxed{\phantom{u}}\right)\) to evaluate the integral in terms of \(\theta\).
\[
\frac{16}{\sqrt{6}} \int \cos^2 \theta \, d\theta = \frac{16}{\sqrt{6}} \int \frac{1}{2} \left(1 + \boxed{\phantom{u}}\right) \, d\theta
\]
\[
= \frac{8}{\sqrt{6}} \left( \boxed{\phantom{u}} \right) + C
\]
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