Use an appropriate substitution and then a trigonometric substitution to evaluate the integral. In 5 S 0 et dt √²+36 Which substitution transforms the given integral into one that can be evaluated directly in terms of 0? OA. et 6 sec 0 OB. et = 6 sin0 C. e-6 tan 0 Find the correct expression for the differential. e¹dt = de

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