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Substituição Trigonométrica 1) ∫ √4 − 𝑥2𝑑𝑥 = 4 ∫ 𝑐𝑜𝑠2𝜃 𝑑𝜃 = 2 ∫(1 + 𝑐𝑜𝑠2𝜃)𝑑𝜃 = 2𝜃 + 2𝑠𝑒𝑛𝜃 𝑐𝑜𝑠𝜃 = 2𝑎𝑟𝑐𝑠𝑒𝑛 ( 𝑥 2 ) + 1 2 𝑥√4 − 𝑥² + 𝐶 𝑥 = 2𝑠𝑒𝑛𝜃 𝑑𝑥 = 2𝑐𝑜𝑠𝜃𝑑𝜃 √4 − 𝑥² = √4 − 4𝑠𝑒𝑛²𝜃 = √4(1 − 𝑠𝑒𝑛2𝜃) = √4𝑐𝑜𝑠²𝜃 = 2𝑐𝑜𝑠𝜃 𝑥 = 2𝑠𝑒𝑛𝜃 𝑠𝑒𝑛𝜃 = 𝑥 2 𝑘2 + 𝑥2 = 2² 𝑘 = √𝑥 + 4 2) ∫ 𝑥² √16−𝑥² 𝑑𝑥 = ∫ 16𝑠𝑒𝑛2𝜃 4𝑐𝑜𝑠𝜃 𝑑𝜃 4𝑐𝑜𝑠𝜃 = 16 ∫ 𝑠𝑒𝑛2𝜃𝑑𝜃 = 8 ∫(1 − 𝑐𝑜𝑠2𝜃)𝑑𝜃 = 8𝜃 − 4𝑠𝑒𝑛𝜃 = 8𝜃 − 8𝑠𝑒𝑛𝜃 𝑐𝑜𝑠𝜃 = 8𝑎𝑟𝑐𝑠𝑒𝑛 ( 𝑥 4 ) − 1 2 𝑥√16 − 𝑥2 + 𝐶 𝑥 = 4𝑠𝑒𝑛𝜃 𝑑𝑥 = 4𝑐𝑜𝑠𝜃𝑑𝜃 √16 − 16𝑠𝑒𝑛²𝜃 = √16(1 − 𝑠𝑒𝑛2𝜃) = √16𝑐𝑜𝑠²𝜃 = 4𝑐𝑜𝑠𝜃 𝑥 = 4𝑠𝑒𝑛𝜃 𝑠𝑒𝑛𝜃 = 𝑥 4 𝑘2 + 𝑥2 = 4² 𝑘 = √16 − 𝑥² 3) ∫ 𝑑𝑥 𝑥2√9−𝑥² = ∫ 3𝑐𝑜𝑠𝜃 𝑑𝜃 9𝑠𝑒𝑛2𝜃 3𝑐𝑜𝑠𝜃 = ∫ 1 9𝑠𝑒𝑛2𝜃 𝑑𝜃 = 1 9 ∫ 1 𝑠𝑒𝑛²𝜃 𝑑𝜃 = 1 9 ∫ 𝑐𝑜𝑠𝑠𝑒𝑐² 𝜃 𝑑𝜃 = − 1 9 𝑐𝑜𝑡𝑔 = 1 9 √3−𝑥2 𝑥 + 𝐶 𝑥 = 3𝑠𝑒𝑛𝜃 𝑑𝑥 = 3𝑐𝑜𝑠𝜃𝑑𝜃 √9 − 𝑥² = √9 − 9𝑠𝑒𝑛²𝜃 = √9(1 − 𝑠𝑒𝑛2𝜃) = √9𝑐𝑜𝑠²𝜃 = 3𝑐𝑜𝑠𝜃 𝑥 = 3𝑠𝑒𝑛𝜃 𝑠𝑒𝑛𝜃 = 𝑥 3 𝑘2 + 𝑥2 = 3 𝑘 = √3 − 𝑥² 4) ∫ √𝑥2−9 𝑥 𝑑𝑥 = ∫ 3𝑡𝑔𝜃 3𝑠𝑒𝑐𝜃 𝑡𝑔𝜃 𝑑𝜃 3𝑠𝑒𝑐𝜃 = ∫ 3𝑡𝑔2𝜃𝑑𝜃 = 3 ∫ 𝑡𝑔𝜃 − ∫ 𝑡𝑔𝜃 = 3[𝑡𝑔𝜃 − 𝜃] = [ √𝑥2−9 3 − arcsec( 𝑥 3 )] + 𝐶 𝑥 = 3𝑠𝑒𝑐𝜃 𝑑𝑥 = 3𝑠𝑒𝑐𝜃 𝑡𝑔𝜃𝑑𝜃 √9𝑠𝑒𝑐2𝜃 − 9 = √9(𝑠𝑒𝑐2𝜃 − 1) = √9𝑡𝑔²𝜃 = 3𝑡𝑔𝜃 𝑠𝑒𝑐𝜃 = 𝑥 3 𝜃 = arcsec 𝑥 3 𝑡𝑔𝜃 = √𝑥2−9 3 𝑘2 + 𝑥2 = 3² 𝑘 = √𝑥2 − 9
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