HEC USAT-E (Pre-Engineering) Mathematics Integration — Set 2

HEC USAT-E (Pre-Engineering) Mathematics Integration — Set 2

1. Question 1

   Q1. ∫1 / (√(x² - 4)) dx = ?

   • A) ln|x + √(x² - 4)| + C
   • B) ln|x - √(x² - 4)| + C
   • C) (1/2)ln|x² - 4| + C
   • D) sin-¹(x/2) + C

   Answer: ln|x + √(x² - 4)| + C

   Explanation: Using standard integral ∫1 / √(x² - a²) dx = ln|x + √(x² - a²)| + C, where a = 2.

2. Question 2

   Q2. ∫e^(2x)sinx dx = ?

   • A) (e^(2x)(2sinx - cosx))/5 + C
   • B) (e^(2x)(sinx + 2cosx))/5 + C
   • C) (e^(2x)(sinx - cosx))/2 + C
   • D) (e^(2x)sinx)/2 + C

   Answer: (e^(2x)(2sinx - cosx))/5 + C

   Explanation: Using integration by parts twice, with u = sinx and dv = e^(2x)dx, we get the required result.

3. Question 3

   Q3. ∫x² / (1 + x³) dx = ?

   • A) (1/3)ln|1 + x³| + C
   • B) (1/2)ln|1 + x²| + C
   • C) (1/3)ln|1 + x| + C
   • D) x - (1/3)ln|1 + x³| + C

   Answer: (1/3)ln|1 + x³| + C

   Explanation: Using substitution u = 1 + x³, du/dx = 3x², hence ∫x² / (1 + x³) dx = (1/3)∫du/u = (1/3)ln|u| + C.

4. Question 4

   Q4. ∫1 / (x² + 4x + 8) dx = ?

   • A) (1/2)tan-¹((x + 2)/2) + C
   • B) (1/4)ln|x² + 4x + 8| + C
   • C) tan-¹(x + 2) + C
   • D) (1/2)ln|x + 2| + C

   Answer: (1/2)tan-¹((x + 2)/2) + C

   Explanation: Completing the square in denominator, x² + 4x + 8 = (x + 2)² + 4, and using standard integral ∫1 / (u² + a²) du.

5. Question 5

   Q5. ∫sin(√x) dx = ?

   • A) 2√xsin(√x) + 2cos(√x) + C
   • B) -2√xcos(√x) + 2sin(√x) + C
   • C) 2sin(√x) - 2√xcos(√x) + C
   • D) √xsin(√x) - cos(√x) + C

   Answer: 2sin(√x) - 2√xcos(√x) + C

   Explanation: Using substitution u = √x, and then integration by parts with dv = sin(u)du.

6. Question 6

   Q6. ∫1 / (1 + sinx) dx = ?

   • A) tanx - secx + C
   • B) tan(π/4 - x/2) + C
   • C) -cot(x/2 + π/4) + C
   • D) secx + tanx + C

   Answer: tanx - secx + C

   Explanation: Using trigonometric identity and substitution to simplify the integral.

7. Question 7

   Q7. ∫e^x / (1 + e^x) dx = ?

   • A) ln|1 + e^x| + C
   • B) e^x - ln|1 + e^x| + C
   • C) ln|e^x| + C
   • D) x + ln|1 + e^x| + C

   Answer: ln|1 + e^x| + C

   Explanation: Using substitution u = 1 + e^x, du/dx = e^x, hence ∫e^x / (1 + e^x) dx = ∫du/u = ln|u| + C.

8. Question 8

   Q8. ∫(sinx + cosx) / (sinx - cosx) dx = ?

   • A) ln|sinx - cosx| + C
   • B) ln|sinx + cosx| + C
   • C) -ln|sinx - cosx| + C
   • D) x + ln|sinx - cosx| + C

   Answer: ln|sinx - cosx| + C

   Explanation: Using substitution u = sinx - cosx, du/dx = cosx + sinx, hence ∫(sinx + cosx) / (sinx - cosx) dx = ∫du/u = ln|u| + C.

9. Question 9

   Q9. ∫1 / (1 + e^x) dx = ?

   • A) x - ln|1 + e^x| + C
   • B) ln|e^x| - ln|1 + e^x| + C
   • C) ln|1 + e^x| + C
   • D) -ln|1 + e^(-x)| + C

   Answer: x - ln|1 + e^x| + C

   Explanation: Using algebraic manipulation and substitution to simplify the integral.

10. Question 10

    Q10. ∫(x + 2) / √(x² + 4x + 7) dx = ?

    • A) √(x² + 4x + 7) + ln|x + 2 + √(x² + 4x + 7)| + C
    • B) √(x² + 4x + 7) + C
    • C) ln|x + 2 + √(x² + 4x + 7)| + C
    • D) √(x² + 4x + 7) - ln|x + 2 + √(x² + 4x + 7)| + C

    Answer: √(x² + 4x + 7) + ln|x + 2 + √(x² + 4x + 7)| + C

    Explanation: Splitting the integral into two parts, and using substitution u = x² + 4x + 7, and standard integral.

11. Question 11

    Q11. ∫√(1 + x²) dx = ?

    • A) (x√(1 + x²) + ln|x + √(1 + x²)|) / 2 + C
    • B) (x√(1 + x²)) / 2 + C
    • C) x√(1 + x²) + C
    • D) √(1 + x²) + C

    Answer: (x√(1 + x²) + ln|x + √(1 + x²)|) / 2 + C

    Explanation: Applied trigonometric substitution x = tan(θ), dx = sec²(θ) dθ, and used trigonometric identities.

12. Question 12

    Q12. ∫e^(x) sin(x) dx = ?

    • A) (e^(x) (sin(x) - cos(x))) / 2 + C
    • B) e^(x) sin(x) + C
    • C) (e^(x) sin(x)) / 2 + C
    • D) (e^(x) cos(x)) / 2 + C

    Answer: (e^(x) (sin(x) - cos(x))) / 2 + C

    Explanation: Applied integration by parts twice, resulting in the formula ∫e^(x) sin(x) dx = (e^(x) (sin(x) - cos(x))) / 2 + C.

13. Question 13

    Q13. ∫cos(x) / (1 + sin(x)) dx = ?

    • A) ln|1 + sin(x)| + C
    • B) ln|sin(x)| + C
    • C) 1 / (1 + sin(x)) + C
    • D) sin(x) + C

    Answer: ln|1 + sin(x)| + C

    Explanation: Used substitution u = 1 + sin(x), du = cos(x) dx, resulting in ∫du / u = ln|u| + C.

14. Question 14

    Q14. ∫1 / (x ln(x)) dx = ?

    • A) ln|ln(x)| + C
    • B) ln|x| + C
    • C) 1 / ln(x) + C
    • D) x ln(x) + C

    Answer: ln|ln(x)| + C

    Explanation: Used substitution u = ln(x), du = 1 / x dx, resulting in ∫du / u = ln|u| + C.

15. Question 15

    Q15. ∫e^(x) (1 + x) / e^(x) dx = ?

    • A) x + C
    • B) x + e^(x) + C
    • C) x + x*e^(x) + C
    • D) x*e^(x) + C

    Answer: x + C

    Explanation: Simplified the integrand to 1 + x, then integrated term by term.

16. Question 16

    Q16. ∫(ln(x))² / x dx = ?

    • A) (1 / 3) (ln(x))³ + C
    • B) (1 / 2) (ln(x))² + C
    • C) ln(x) + C
    • D) (ln(x))³ + C

    Answer: (1 / 3) (ln(x))³ + C

    Explanation: Used substitution u = ln(x), du = 1 / x dx, resulting in ∫u² du = (1 / 3)u³ + C.

17. Question 17

    Q17. ∫cos(x) / √(sin(x)) dx = ?

    • A) 2√(sin(x)) + C
    • B) √(sin(x)) + C
    • C) (1 / 2) √(sin(x)) + C
    • D) sin(x) + C

    Answer: 2√(sin(x)) + C

    Explanation: Used substitution u = sin(x), du = cos(x) dx, resulting in ∫du / √u = 2√u + C.

18. Question 18

    Q18. ∫(sin²x + cos²x) dx from 0 to π/2 = ?

    • A) π/4
    • B) π/2
    • C) 1
    • D) π

    Answer: π/2

    Explanation: Since sin²x + cos²x = 1, the integral becomes ∫1 dx from 0 to π/2, which equals π/2.

19. Question 19

    Q19. ∫x² sin(x) dx = ?

    • A) -x²cos(x) + 2∫xcos(x) dx
    • B) x²cos(x) + 2∫xcos(x) dx
    • C) -x²sin(x) + 2∫xsin(x) dx
    • D) x²sin(x) + 2∫xsin(x) dx

    Answer: -x²cos(x) + 2∫xcos(x) dx

    Explanation: Using integration by parts with u = x² and dv = sin(x) dx, we get du = 2x dx and v = -cos(x).

20. Question 20

    Q20. ∫(2x) / (x² + 1) dx = ?

    • A) ln|x² + 1| + C
    • B) ln|x + 1| + C
    • C) (1/2)ln|x² + 1| + C
    • D) x + ln|x² + 1| + C

    Answer: ln|x² + 1| + C

    Explanation: Using substitution u = x² + 1, du/dx = 2x, the integral simplifies to ∫du/u = ln|u| + C.