27T SES 22 sinlu) I+4cos²u dudv

Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter2: Working With Real Numbers
Section2.9: Dividing Real Numbers
Problem 9MRE
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Question
217
SS2sinlu)+4cosu du dv
+4(05²u dudu
Transcribed Image Text:217 SS2sinlu)+4cosu du dv +4(05²u dudu
Expert Solution
Step 1

Given integral is,

0702π22sinx 1+4cos2xdxdy

Step 2

To find 0702π22sinx 1+4cos2xdxdy.

0702π22sinx 1+4cos2xdxdy

The integral of 02π22sinx1+4cos2xdx=-2cosx4cos2x+1+12ln2cosx+1+4cos2x02π                                                    =-2cos2π4cos22π+1+12ln2cos2π+1+4cos22π--2cos04cos20+1+12ln2cos0+1+4cos20                                                    =-2(5)+12ln2(1)+5+25-12ln2+5                                                    =0

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