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
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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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 \(\theta\).

\[
\frac{16}{\sqrt{6}} \int \cos^2 \theta \, d\theta = \frac{16}{\sqrt{6}} \int \frac{1}{2} \left(1 + \boxed{\phantom{u}}\right) \, d\theta
\]

\[
= \frac{8}{\sqrt{6}} \left( \boxed{\phantom{u}} \right) + C
\]
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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