Sec& sec'x + secikm 8.3. Trigonometric Integrals 721 tan 0 sec" 0 do = 8.3.38 (tan 0) (tan 0+ 1)(sec2 0) d0. Let u = tan 0 so that du = sec2 0 d0. Substi- tuting yields u+u° du = +C = 8. tans 0 tan 0 +C. u=cosx du = - sinx 8. Sinx 8.3.39 COSX
Sec& sec'x + secikm 8.3. Trigonometric Integrals 721 tan 0 sec" 0 do = 8.3.38 (tan 0) (tan 0+ 1)(sec2 0) d0. Let u = tan 0 so that du = sec2 0 d0. Substi- tuting yields u+u° du = +C = 8. tans 0 tan 0 +C. u=cosx du = - sinx 8. Sinx 8.3.39 COSX
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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The answer the answer key provided is circled in yellow highlighter. My question is, could the part circled in pink also be an answer? Where it says sec^8x/8-1/3 sec^6x + sec^4x/4 +C? That is what I got.
Transcribed Image Text:Sec&
X + secyx
3
6
sec
8.3. Trigonometric Integrals
721
8.3.38
tan e sec* e de
(tan 0) (tan? 0 + 1)(sec² 0) d0. Let u =
%3D
tan 0 so that du
SO
sec2 0 d0. Substi-
tan 0
8
u=COSX
du=-sinx
tuting yields
u +u° du
8.
+C =
6.
tan6 0
+C.
6.
8.
sinx
8.3.39
COSX
8.
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