engineering mathematics MCQ #1533

∫(sin³x) / (cosx) dx = ?

engineering mathematics MCQ #1533

  1. Question 1

    Q1. ∫(sin³x) / (cosx) dx = ?

    • A) (1/2)sin²x - ln|cosx| + C
    • B) (1/3)sin³x + C
    • C) -cosx + (1/3)cos³x + C
    • D) -(1/2)cos²x + ln|sinx| + C

    Answer: (1/2)sin²x - ln|cosx| + C

    Explanation: Using substitution u = cosx, du/dx = -sinx, and sin²x = 1 - cos²x, hence ∫(sin³x) / (cosx) dx = -∫(1 - u²)/u du.