Question: Question 7: (1 point) The goal of this question is to guide you through the evaluation of the following integral, using an appropriate trigonometric

Question 7: (1 point) The goal of this question is to guideyou through the evaluation of the following integral, using an appropriate trigonometric

substitution: a) First, select one of the following trig identities to bethe backbone of your substitution: 1 sin2(g) = cos2 (9) sec2 (9)

Question 7: (1 point) The goal of this question is to guide you through the evaluation of the following integral, using an appropriate trigonometric substitution: a) First, select one of the following trig identities to be the backbone of your substitution: 1 sin2(g) = cos2 (9) sec2 (9) 1 = tan2(9) tan2(9) + 1 = see (9) 64 Hint: Choose the trigonometric identity whose left side most resembles z2 The right side of each identity is a perfect square which will ultimately help you to eliminate the radical and simplify the integral! b) Next, factor out 64 from the radical to obtain an expression of the form 64 = 8 .42 1 What is A? FORMAT de la RPONSE: La formule pour A va dpende de Puisque /'on ne peut pas crire /8 lettre Grecque 9 ('meta") dans la rponse pour Mbius, nous utiljserons 'a variable Q la place de e pour les pallies qui restent de la question c) Choisir la substitution pour x pour que rexpression .42 1 (en fonction de x) soit exactementle ct gauche de l'identit trigonomtrique en (a). 8tan2(Q) = 8sin2(Q) 8sec2 (Q) sec(Q) 8 J tan(Q) 8 = 8sec(Q) sin(Q) 8 = 8tan(Q) 8 sin(Q) (where (where 2 (where 0 S Q < (where 0 Q < (where 2 (where 0 Q < (where where (where 2 or < Q or < Q 2 Ir or -L < Q d) partir de la substitution en crire Q en fonction de x raide de la fonction trigonomtrique inverse approprie_ FORMAT de la REPONSE: Portez attention /a syntaxe utilise dans Mbius pow dnoter jes fonct\ons trigonomtriques inverses; e.g. sin (a) est crit arcsec(a), et tanl (a) est crjt arctan(a) , etc (a) est cr,it arcsin(a), sec

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