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Integrals Involving Inverse Trigonometric Functions quiz

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  • What is the integral of 1 over the square root of 1 minus x squared dx?
    The integral is the inverse sine of x plus a constant, written as sin⁻¹(x) + C.
  • How do you integrate 4 over the square root of 1 minus x squared dx?
    Pull out the constant to get 4 times the integral, resulting in 4 sin⁻¹(x) + C.
  • What is the integral of 1 over 1 plus x squared dx?
    The integral is the inverse tangent of x plus a constant, tan⁻¹(x) + C.
  • How do you integrate 1 over x squared dx?
    Use the power rule to get -1/x plus a constant, -1/x + C.
  • What is the integral of 1 over the absolute value of x times the square root of x squared minus 1 dx?
    The integral is the inverse secant of x plus a constant, sec⁻¹(x) + C.
  • How do you integrate 1 over 3 times the absolute value of x times the square root of x squared minus 1 dx?
    Pull out the constant to get 1/3 times the integral, resulting in (1/3) sec⁻¹(x) + C.
  • What is the integral of 1 over the square root of a squared minus u squared du?
    The integral is sin⁻¹(u/a) + C, where a is a positive constant and u is a function of x.
  • How do you integrate 1 over the square root of 16 minus 9x squared dx?
    Rewrite as 1 over the square root of 4² minus (3x)², substitute u = 3x, and get (1/3) sin⁻¹(3x/4) + C.
  • What is the integral of 1 over a squared plus u squared du?
    The integral is (1/a) tan⁻¹(u/a) + C.
  • How do you integrate 6 over 25 plus 36x squared dx?
    Rewrite as 6 over 5² plus (6x)², substitute u = 6x, and get (1/5) tan⁻¹(6x/5) + C.
  • What is the integral of 1 over the absolute value of u times the square root of u squared minus a squared du?
    The integral is (1/a) sec⁻¹(|u|/a) + C.
  • How do you integrate 7 over the absolute value of 7x times the square root of 49x squared minus 4 dx?
    Rewrite as 7 over |7x| times sqrt((7x)² - 2²), substitute u = 7x, and get (1/2) sec⁻¹(|7x|/2) + C.
  • What technique can be used to rewrite a quadratic denominator to match an inverse trig integral form?
    Completing the square can transform the denominator into a recognizable form for inverse trig integrals.
  • How do you complete the square for x² + 6x + 13?
    Add and subtract (6/2)² = 9 to get x² + 6x + 9 + 13 - 9 = (x + 3)² + 4.
  • How do you integrate 7 over x² + 6x + 13 dx using inverse trig formulas?
    Complete the square to get 7 over (x + 3)² + 2², then use the inverse tangent rule to get (7/2) tan⁻¹((x + 3)/2) + C.