Find 1 x²√x² + 16 J Solution Let x = 4 tan(0), - Thus, we have sec(0) tan²(0) dx. -<6 2 = << . Then dx 2 √x² + 16 = √16 (tan² (0) + 1) = √ 16 sec²(0) = 4|sec(0)| = 4 sec(0). dx [x²√x² +16 = 1 16 = ᏧᎾ 16 tan2(0) 4 sec(0) . cos²(8) sin²(0) tan²(0) To evaluate this trigonometric integral we put everything in terms of sin(0) and cos(0). 1 de and ᏧᎾ .

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