Problem 6. PREVIEW ONLY -- ANSWERS NOT RECORDED (1 point) Trigonometric Substitutions The three basic trigonometric...

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Problem 6. PREVIEW ONLY -- ANSWERS NOT RECORDED (1 point) Trigonometric Substitutions The three basic trigonometric substitutions are in the table below. Part 1. Part 2. Using the substitution: a = 5 sec 0,0 <0, re-write the indefinite integral then evaluate in terms of 0. Suppose that > 5. Then, 1 √24√/2²2² =25° dx =S₁ Note: type 'theta' for 0 and 'dtheta' for de. dx Integral Contains √a²-x² √a² + x² √x² a² 1 Back substituting in the antiderivative you found in Part 1. above we have 1 √2+√7²-25 = Try using Substitution x = a sin 0, x = a tan 0, x = a sec 0, || ⠀ 2 << < < 2 0<0< KIN 프 2 or << T Problem 6. PREVIEW ONLY -- ANSWERS NOT RECORDED (1 point) Trigonometric Substitutions The three basic trigonometric substitutions are in the table below. Part 1. Part 2. Using the substitution: a = 5 sec 0,0 <0, re-write the indefinite integral then evaluate in terms of 0. Suppose that > 5. Then, 1 √24√/2²2² =25° dx =S₁ Note: type 'theta' for 0 and 'dtheta' for de. dx Integral Contains √a²-x² √a² + x² √x² a² 1 Back substituting in the antiderivative you found in Part 1. above we have 1 √2+√7²-25 = Try using Substitution x = a sin 0, x = a tan 0, x = a sec 0, || ⠀ 2 << < < 2 0<0< KIN 프 2 or << T

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EQUATIONS USED sectan 1 sec 1 tan0 sin cos 1 dx de dx d de sece tane Given sec 1 Put x 4... View the full answer