The following integral suggests trigonometric substitution.x = 4 sin(0); 2 J da. (16-x)/2 After making this...

 

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The following integral suggests trigonometric substitution.x = 4 sin(0); 2² J₁ da. (16-x²)³/2 After making this substitution and integrating, we obtain J 2² (16-x²)³/2 = dx = [ ²=² = sin² (0) cos(0) (cos² (6))3/2 do = [ tan² (0) do = [sec¹²( sec² (0) 1 de tan(0) - 0 + C. = The final step is to return to the original variable z. Fill in the given right triangle and then write the antiderivative in terms of z. Do NOT use sine, cosine, or tangent functions in your final answer (but you may have inverse trigonometric functions in your answer). The following integral suggests trigonometric substitution.x = 4 sin(0); 2² J₁ da. (16-x²)³/2 After making this substitution and integrating, we obtain J 2² (16-x²)³/2 = dx = [ ²=² = sin² (0) cos(0) (cos² (6))3/2 do = [ tan² (0) do = [sec¹²( sec² (0) 1 de tan(0) - 0 + C. = The final step is to return to the original variable z. Fill in the given right triangle and then write the antiderivative in terms of x. Do NOT use sine, cosine, or tangent functions in your final answer (but you may have inverse trigonometric functions in your answer). The following integral suggests trigonometric substitution.x = 4 sin(0); 2² J₁ da. (16-x²)³/2 After making this substitution and integrating, we obtain J 2² (16-x²)³/2 = dx = [ ²=² = sin² (0) cos(0) (cos² (6))3/2 do = [ tan² (0) do = [sec¹²( sec² (0) 1 de tan(0) - 0 + C. = The final step is to return to the original variable z. Fill in the given right triangle and then write the antiderivative in terms of z. Do NOT use sine, cosine, or tangent functions in your final answer (but you may have inverse trigonometric functions in your answer). The following integral suggests trigonometric substitution.x = 4 sin(0); 2² J₁ da. (16-x²)³/2 After making this substitution and integrating, we obtain J 2² (16-x²)³/2 = dx = [ ²=² = sin² (0) cos(0) (cos² (6))3/2 do = [ tan² (0) do = [sec¹²( sec² (0) 1 de tan(0) - 0 + C. = The final step is to return to the original variable z. Fill in the given right triangle and then write the antiderivative in terms of x. Do NOT use sine, cosine, or tangent functions in your final answer (but you may have inverse trigonometric functions in your answer).