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MATH 142 EXAM 200 Practice Questions with Verified Answers & Explanations (2026 Edition) SECTION 1: INTEGRATION TECHNIQUES (Questions 1–50)
Subsection 1.1: Basic Integration (Questions 1–15)
- Evaluate ∫ x² e^(-x³) dx A) - ⅓ e^(-x³) + C B) ⅓ e^(-x³) + C C) - x³ e^(-x³) + C D) e^(-x³) + C Answer: A Explanation: Using substitution u = - x³, du = - 3x² dx. Then x² dx = - ⅓ du. The integral becomes - ⅓ ∫ e^u du = - ⅓ e^u + C = - ⅓ e^(-x³) + C.
- Evaluate ∫₁^e (√(ln x))/x dx A) 1 B) ⅔
C) 2/
D) e Answer: C Explanation: Let u = ln x, du = (1/x) dx. When x = 1, u = 0; when x = e, u = 1. The integral becomes ∫₀¹ √u du = [⅔ u^(3/2)]₀¹ = ⅔.
- Evaluate ∫ (sin t)/(cos t + 1)² dt A) 1/(cos t + 1) + C B) - 1/(cos t + 1) + C C) ln|cos t + 1| + C D) sec² t + C Answer: A Explanation: Let u = cos t + 1, du = - sin t dt. The integral becomes - ∫ u^(-2) du = 1/u + C = 1/(cos t + 1) + C.
- Evaluate ∫ x (x² + 1)^(3/2) dx A) ⅕(x² + 1)^(5/2) + C
A) x³ + x² + x + C B) x³ + x² + x C) 3x³ + 2x² + x + C D) x³ + 2x² + x + C Answer: A Explanation: ∫ 3x² dx = x³, ∫ 2x dx = x², ∫ 1 dx = x. Sum = x³ + x² + x + C.
- Evaluate ∫ (1/x) dx A) ln|x| + C B) ln x + C C) 1/x² + C D) x ln x + C Answer: A Explanation: The integral of 1/x is ln|x| + C.
- Evaluate ∫ e^(2x) dx
A) 2e^(2x) + C B) ½ e^(2x) + C C) e^(2x) + C D) e^(2x)/ Answer: B Explanation: ∫ e^(kx) dx = (1/k)e^(kx) + C. Here k = 2, so ∫ e^(2x) dx = ½ e^(2x) + C.
- Evaluate ∫ sin(3x) dx A) - 3 cos(3x) + C B) - ⅓ cos(3x) + C C) ⅓ cos(3x) + C D) 3 cos(3x) + C Answer: B Explanation: ∫ sin(kx) dx = - (1/k)cos(kx) + C. Here k = 3, so ∫ sin(3x) dx = - ⅓ cos(3x) + C.
- Evaluate ∫ cos(2x) dx
A) (2x + 3)⁶/12 + C B) (2x + 3)⁶/6 + C C) (2x + 3)⁶/3 + C D) (2x + 3)⁶/2 + C Answer: A Explanation: Let u = 2x + 3, du = 2 dx. The integral becomes ½ ∫ u⁵ du = ½ · u⁶/6 + C = (2x + 3)⁶/
- Evaluate ∫ (x² + 1)³ x dx A) (x² + 1)⁴/8 + C B) (x² + 1)⁴/4 + C C) (x² + 1)⁴ + C D) (x² + 1)⁴/2 + C Answer: A Explanation: Let u = x² + 1, du = 2x dx, so x dx = ½ du. The integral becomes ½ ∫ u³ du = ½ · u⁴/4 + C = (x² + 1)⁴/8 + C.
- Evaluate ∫₁² (2x + 1) dx
A) 2
B) 3
C) 4
D) 5
Answer: C Explanation: ∫₁² (2x + 1) dx = [x² + x]₁² = (4 + 2) - (1 + 1) = 6 - 2 = 4.
- Evaluate ∫₀^π sin x dx A) 0 B) 1 C) 2 D) π Answer: C Explanation: ∫₀^π sin x dx = [-cos x]₀^π = - cos π - (-cos 0) = - (-1) - (-1) = 1 + 1 = 2.
Subsection 1.2: Integration by Parts (Questions 16–30)
- Evaluate ∫ x cos x dx A) x sin x + cos x + C B) x sin x - cos x + C C) - x sin x + cos x + C D) x cos x + sin x + C Answer: A Explanation: Let u = x, dv = cos x dx. Then du = dx, v = sin x. ∫ x cos x dx = x sin x - ∫ sin x dx = x sin x + cos x + C.
- Evaluate ∫ ln x dx A) x ln x - x + C B) x ln x + x + C C) ln x - x + C D) x ln x + C Answer: A Explanation: Let u = ln x, dv = dx. Then du = (1/x) dx, v = x. ∫ ln x dx = x ln x - ∫ 1 dx = x ln x - x + C.
- Evaluate ∫ x ln x dx A) ½ x² ln x - ¼ x² + C B) ½ x² ln x + ¼ x² + C C) x² ln x - ¼ x² + C D) ½ x² ln x - ½ x² + C Answer: A Explanation: Let u = ln x, dv = x dx. Then du = (1/x) dx, v = ½ x². ∫ x ln x dx = ½ x² ln x - ∫ ½ x dx = ½ x² ln x - ¼ x² + C.
- Evaluate ∫ x² e^x dx A) e^x(x² - 2x + 2) + C B) e^x(x² + 2x + 2) + C C) e^x(x² - 2x) + C D) e^x(x² + 2x) + C Answer: A
Explanation: Let u = x, dv = sec² x dx. Then du = dx, v = tan x. ∫ x sec² x dx = x tan x - ∫ tan x dx = x tan x - (-ln|cos x|) + C = x tan x + ln|cos x| + C.
- Evaluate ∫ e^x cos x dx A) ½ e^x(cos x + sin x) + C B) ½ e^x(cos x - sin x) + C C) e^x(cos x + sin x) + C D) ½ e^x(sin x - cos x) + C Answer: A Explanation: Using integration by parts twice: Let I = ∫ e^x cos x dx. I = e^x sin x - ∫ e^x sin x dx = e^x sin x + e^x cos x - I. Therefore, 2I = e^x(sin x + cos x), so I = ½ e^x(cos x + sin x) + C.
- Evaluate ∫ e^x sin x dx A) ½ e^x(sin x - cos x) + C B) ½ e^x(sin x + cos x) + C C) e^x(sin x - cos x) + C D) ½ e^x(cos x - sin x) + C
Answer: A Explanation: Using integration by parts twice: Let I = ∫ e^x sin x dx. I = - e^x cos x + ∫ e^x cos x dx = - e^x cos x + e^x sin x - I. Therefore, 2I = e^x(sin x - cos x), so I = ½ e^x(sin x - cos x) + C.
- Evaluate ∫ x² ln x dx A) ⅓ x³ ln x - ¹/₉ x³ + C B) ⅓ x³ ln x + ¹/₉ x³ + C C) x³ ln x - ⅓ x³ + C D) ⅓ x³ ln x - ⅓ x³ + C Answer: A Explanation: Let u = ln x, dv = x² dx. Then du = (1/x) dx, v = ⅓ x³. ∫ x² ln x dx = ⅓ x³ ln x - ⅓ ∫ x² dx = ⅓ x³ ln x - ¹/₉ x³ + C.
- Evaluate ∫ x arctan x dx A) ½ x² arctan x - ½ x + ½ arctan x + C B) ½ x² arctan x + ½ x - ½ arctan x + C C) ½ x² arctan x - ½ x - ½ arctan x + C D) x² arctan x - x + arctan x + C
C) x² cos x + 2x sin x - 2 cos x + C D) - x² cos x - 2x sin x + 2 cos x + C Answer: A Explanation: Using integration by parts twice: Let u = x², dv = sin x dx. Then du = 2x dx, v = - cos x. ∫ x² sin x dx = - x² cos x + 2∫ x cos x dx = - x² cos x + 2(x sin x + cos x) + C = - x² cos x + 2x sin x + 2 cos x + C.
- Evaluate ∫ arctan x dx A) x arctan x - ½ ln(1 + x²) + C B) x arctan x + ½ ln(1 + x²) + C C) x arctan x - ln(1 + x²) + C D) arctan x - ½ ln(1 + x²) + C Answer: A Explanation: Let u = arctan x, dv = dx. Then du = 1/(1+x²) dx, v = x. ∫ arctan x dx = x arctan x - ∫ x/(1+x²) dx = x arctan x - ½ ln(1 + x²) + C.
Subsection 1.3: Trigonometric Integrals (Questions 31–40)
- Evaluate ∫ sin² x dx A) ½ x - ¼ sin(2x) + C B) ½ x + ¼ sin(2x) + C C) ½ x - ½ sin(2x) + C D) x - ½ sin(2x) + C Answer: A Explanation: Using sin² x = (1 - cos(2x))/2. ∫ sin² x dx = ½∫(1 - cos(2x)) dx = ½ x - ¼ sin(2x) + C.
- Evaluate ∫ cos² x dx A) ½ x + ¼ sin(2x) + C B) ½ x - ¼ sin(2x) + C C) ½ x + ½ sin(2x) + C D) x + ¼ sin(2x) + C Answer: A Explanation: Using cos² x = (1 + cos(2x))/2. ∫ cos² x dx = ½∫(1 + cos(2x)) dx = ½ x + ¼ sin(2x) + C.
- Evaluate ∫ sin⁴ x dx A) ⅜ x - ¼ sin(2x) + ¹/₃₂ sin(4x) + C B) ⅜ x - ¼ sin(2x) + ¹/₃₂ sin(4x) + C C) ⅜ x + ¼ sin(2x) + ¹/₃₂ sin(4x) + C D) ⅜ x - ½ sin(2x) + ¹/₃₂ sin(4x) + C Answer: B Explanation: sin⁴ x = (sin² x)² = ((1 - cos(2x))/2)² = ¼(1 - 2cos(2x) + cos²(2x)) = ¼(1 - 2cos(2x) + (
- cos(4x))/2) = ⅜ - ½ cos(2x) + ⅛ cos(4x). Integrating: ⅜ x - ¼ sin(2x) + ¹/₃₂ sin(4x) + C.
- Evaluate ∫ tan x dx A) ln|sec x| + C B) - ln|cos x| + C C) Both A and B D) ln|sin x| + C Answer: C Explanation: ∫ tan x dx = ∫ sin x/cos x dx = - ln|cos x| + C = ln|sec x| + C. Both A and B are equivalent.
- Evaluate ∫ sec x dx A) ln|sec x + tan x| + C B) ln|sec x - tan x| + C C) ln|tan x| + C D) sec x tan x + C Answer: A Explanation: ∫ sec x dx = ln|sec x + tan x| + C.
- Evaluate ∫ tan² x dx A) tan x - x + C B) tan x + x + C C) sec² x + C D) - tan x - x + C Answer: A Explanation: tan² x = sec² x - 1. ∫ tan² x dx = ∫ (sec² x - 1) dx = tan x - x + C.
