Questions & explanations
1. The integral ∫ sec x/(1 + csc x) dx simplifies to which of the following?
- tan x - sec x + C
- tan x + sec x + C
- sec x - tan x + C
- cot x - csc x + C
Answer: cot x - csc x + C
Rewrite integrand as sin x/(cos x(1+sin x)). Multiply numerator and denominator by (1-sin x) to get sin x(1-sin x)/cos^3 x = tan x sec^2 x - sin^2 x/cos^3 x. Integrate: ∫ tan x sec^2 x dx = (1/2) tan^2 x, and ∫ sin^2 x/cos^3 x dx = ∫ sec^3 x dx - ∫ sec x dx = (1/2)(sec x tan x + ln|sec x+tan x|) - ln|sec x+tan x| = (1/2)(sec x tan x - ln|sec x+tan x|). Combining yields cot x - csc x + C.
2. Evaluate ∫ e^x sin x dx.
- (e^x/2)(sin x + cos x) + C
- (e^x/2)(sin x - cos x) + C
- e^x(sin x - cos x) + C
- (e^x/2)(cos x - sin x) + C
Answer: (e^x/2)(sin x - cos x) + C
Let I = ∫ e^x sin x dx. Apply integration by parts twice: first with u = sin x, dv = e^x dx → I = e^x sin x - ∫ e^x cos x dx. Then on J = ∫ e^x cos x dx with u = cos x, dv = e^x dx → J = e^x cos x + I. Substituting gives I = e^x sin x - (e^x cos x + I) → 2I = e^x(sin x - cos x) → I = (e^x/2)(sin x - cos x) + C.
3. Evaluate ∫_0^π x sin x / (1 + cos^2 x) dx.
- π^2/2
- π^2/4
- π/2
- π^2/8
Answer: π^2/4
Use property P3: ∫_0^π x f(sin x) dx = (π/2) ∫_0^π f(sin x) dx. Here f(sin x) = sin x/(1+cos^2 x). So I = (π/2) ∫_0^π sin x/(1+cos^2 x) dx. Substitute u = cos x, du = -sin x dx, limits: 1 to -1. Integral becomes (π/2) ∫_{-1}^1 du/(1+u^2) = (π/2)[arctan u]_{-1}^1 = (π/2)(π/4 - (-π/4)) = (π/2)(π/2) = π^2/4.
4. If ∫(3x+5)/[(x-1)(x+2)(x-3)] dx = A ln|x-1| + B ln|x+2| + C ln|x-3| + K, then A+B+C equals:
- 1
- 0
- 2
- 3
Answer: 0
Using partial fractions, (3x+5)/[(x-1)(x+2)(x-3)] = A/(x-1) + B/(x+2) + C/(x-3). By cover-up method: A = (3(1)+5)/[(1+2)(1-3)] = 8/(3*(-2)) = -4/3; B = (3(-2)+5)/[(-2-1)(-2-3)] = (-1)/((-3)*(-5)) = -1/15; C = (3(3)+5)/[(3-1)(3+2)] = 14/(2*5) = 7/5. Sum A+B+C = -4/3 - 1/15 + 7/5 = (-20 -1 +21)/15 = 0.
5. What is the value of ∫_0^{π/2} e^x sin x dx?
- e^{π/2} + 1
- (e^{π/2} - 1)/2
- (e^{π/2} + 1)/2
- e^{π/2} - 1
Answer: (e^{π/2} + 1)/2
Let I = ∫_0^{π/2} e^x sin x dx. Use integration by parts twice: first with u=sin x, dv=e^x dx gives I = e^{π/2} - J where J = ∫ e^x cos x dx. Then parts on J: u=cos x, dv=e^x dx gives J = -1 + I. Substituting, I = e^{π/2} + 1 - I, so 2I = e^{π/2} + 1, hence I = (e^{π/2} + 1)/2.
6. Evaluate ∫_0^{π/4} ln(1 + tan x) dx.
- (π/8) ln 2
- (π/4) ln 2
- (π/2) ln 2
- (π/8) ln 4
Answer: (π/8) ln 2
Use property P4: ∫_0^{π/4} f(x) dx = ∫_0^{π/4} f(π/4 - x) dx. Then I = ∫_0^{π/4} ln(1+tan(π/4-x)) dx. Using tan(π/4-x) = (1-tan x)/(1+tan x), we get 1+tan(π/4-x) = 2/(1+tan x). So I = ∫_0^{π/4} [ln 2 - ln(1+tan x)] dx = (π/4) ln 2 - I, giving 2I = (π/4) ln 2, so I = (π/8) ln 2.
7. Evaluate ∫ dx / (5 + 4 sin x) using t = tan(x/2).
- (1/3) tan⁻¹((5 tan(x/2) + 4)/3) + C
- (2/3) tan⁻¹((5 tan(x/2) - 4)/3) + C
- (2/3) tan⁻¹((5 tan(x/2) + 4)/5) + C
- (2/3) tan⁻¹((5 tan(x/2) + 4)/3) + C
Answer: (2/3) tan⁻¹((5 tan(x/2) + 4)/3) + C
Substitute sin x = 2t/(1+t^2), dx = 2 dt/(1+t^2). Then integral becomes ∫ 2 dt / (5(1+t^2) + 8t) = ∫ 2 dt / (5t^2 + 8t + 5). Complete square: 5(t+4/5)^2 + 9/5. So integral = ∫ 2 dt / [5(t+4/5)^2 + 9/5] = (2/3) tan⁻¹((5t+4)/3) + C = (2/3) tan⁻¹((5 tan(x/2)+4)/3) + C.
8. Evaluate ∫_0^π x sin x / (1 + cos^2 x) dx.
- π^2/8
- π^2/2
- π/2
- π^2/4
Answer: π^2/4
Use property P3: ∫_0^π f(x) dx = ∫_0^π f(π-x) dx. Then I = ∫_0^π (π-x) sin x/(1+cos^2 x) dx. Adding gives 2I = π ∫_0^π sin x/(1+cos^2 x) dx. Substitute u = cos x, du = -sin x dx, limits 1 to -1, giving 2I = π ∫_{-1}^1 du/(1+u^2) = π * (π/2) = π^2/2, so I = π^2/4.
9. Evaluate lim_{n→∞} (1/n) Σ_{r=1}^n sin(rπ/n) / (sin(rπ/n) + cos(rπ/n)).
- π/2
- π/4
- 1
- 1/2
Answer: 1/2
The limit equals ∫_0^1 sin(πx)/(sin(πx)+cos(πx)) dx. Substitute u=πx, dx=du/π, limits 0 to π: (1/π)∫_0^π sin u/(sin u+cos u) du. Use property P3: replace u by π-u, add the two forms to get 2J = ∫_0^π du = π, so J = π/2. Thus the integral = (1/π)(π/2) = 1/2.
10. Evaluate ∫ x arctan(x^2) dx.
- (1/2) x^2 arctan(x^2) + (1/4) ln(1+x^4) + C
- x^2 arctan(x^2) - (1/2) ln(1+x^4) + C
- (1/2) x^2 arctan(x^2) - (1/2) ln(1+x^4) + C
- (1/2) x^2 arctan(x^2) - (1/4) ln(1+x^4) + C
Answer: (1/2) x^2 arctan(x^2) - (1/4) ln(1+x^4) + C
Substitute u = x^2, du = 2x dx, so x dx = du/2. Integral becomes (1/2)∫ arctan u du. Integration by parts: ∫ arctan u du = u arctan u - (1/2) ln(1+u^2) + C. Multiply by 1/2: (1/2)u arctan u - (1/4) ln(1+u^2) + C. Substitute back u = x^2 gives option a.
11. Evaluate ∫ (2x+3)/((x^2+1)(x^2+4)) dx.
- (1/3) ln|x^2+1| + (1/2) arctan(x/2) - (1/3) ln|x^2+4| - (1/2) arctan(x) + C
- (1/3) ln|x^2+1| + (1/2) arctan(x) - (1/3) ln|x^2+4| + (1/2) arctan(x/2) + C
- (1/3) ln|x^2+1| + arctan(x) - (1/3) ln|x^2+4| - (1/2) arctan(x/2) + C
- (1/3) ln|x^2+1| - (1/2) arctan(x) - (1/3) ln|x^2+4| + (1/2) arctan(x/2) + C
Answer: (1/3) ln|x^2+1| + arctan(x) - (1/3) ln|x^2+4| - (1/2) arctan(x/2) + C
Partial fractions: (2x+3)/((x^2+1)(x^2+4)) = (2x/3+1)/(x^2+1) + (-2x/3-1)/(x^2+4). Integrate: ∫(2x/3)/(x^2+1)dx = (1/3)ln|x^2+1|; ∫1/(x^2+1)dx = arctan x; ∫(-2x/3)/(x^2+4)dx = -(1/3)ln|x^2+4|; ∫-1/(x^2+4)dx = -(1/2)arctan(x/2). Sum gives option c.
12. Evaluate ∫ (x³ + 2x² + 1) / (x² - 1) dx.
- x²/2 + 2x + 2 ln|x-1| - ln|x+1| + C
- x²/2 + 2x + ln|x-1| + 2 ln|x+1| + C
- x²/2 + 2x + ln|x-1| - 2 ln|x+1| + C
- x²/2 + 2x + 2 ln|x-1| + ln|x+1| + C
Answer: x²/2 + 2x + 2 ln|x-1| - ln|x+1| + C
Using polynomial division: (x³+2x²+1) ÷ (x²-1) = x+2 + (x+3)/(x²-1). Decompose (x+3)/((x-1)(x+1)) = 2/(x-1) - 1/(x+1). Integrate termwise: ∫(x+2)dx = x²/2+2x, ∫2/(x-1)dx = 2ln|x-1|, ∫-1/(x+1)dx = -ln|x+1|. Combine constants to get option d.