Question 1 to Question 6, thank you!

Questions For each of the following problems, always start with drawing an appropriate right triangle for trigonometric substitution. 1. Set up, but do not evaluate, an appropriate trigonometric integral for the following integrals. (a) 1+ 12 = dr (b) V9 - 12 (c) (9 - 472)3/2 dr MTH 200 Trigonometric Substitution - Page 2 of 2 HW 07 2. We will evaluate VI - 1 dr in the following steps. (a) Transform the above integral with r = sin 0, draw the triangle. (b) Use sin?0 = 1 - cos' 0 to integrate your result in part (a). 3. We will evaluate VE - ar in the following steps. (a) Using an appropriate trigonometric substitution, transform the above integral into sin? @ cos 0 de. (b) With ordinary u-substitution technique on part (a), show that ( Viz - dy = 1 (12 - 1)3/2 3 + C. 4. Evaluate VI+r dax with ordinary u-substitution technique. 5. We will evaluate Vi+ 2 dr in the following steps. I (a) Using an appropriate trigonometric substitution, transform the above integral into csc 0 sec2 0 de. (b) Verify the following trigonometric identity: csc 0 sec2 0 = csco + tan 0 sec 0. (c) Using the results in parts (a) and (b), plus the antidervative formlae, show that Vi+ 12 - dr = - In 6. We will evaluate VI2 - dr in the following steps. (a) Using an appropriate trigonometric substitution, transform the above integral into 5 / sin ' 0 de. (b) Integrate sin?0 do using the Half-Angle formula. (c) Using the results in parts (a) and (b), plus the Double-Angle formula, show that VI - 9 73 dx = = sec-1 (V+ -9 + C. 212