Evaluate by using first substitution and then partial fractions if necessary. sec2 d tan01 S
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Evaluate by using first substitution and then partial fractions if necessary.
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sec2 Ꮎ dᎾ tan²01 S
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Let utan 0 then du sec 0 de and sec 0 de tan01 Clearing denominators gives us u The partia...View the full answer
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