🧮 Unit 3 – Part A (2-Mark Q&A)

Sol: Here, n=13 is odd.
∫₀^π/2 cosⁿx dx = (n-1)/n · (n-3)/(n-2) · (n-5)/(n-4) ··· 2/3
∫₀^π/2 cos¹³x dx = (12/13) · (10/11) · (8/9) · (6/7) · (4/5) · (2/3)
= (12 · 10 · 8 · 6 · 4 · 2) / (13 · 11 · 9 · 7 · 5 · 3 · 1)
= (4 · 2 · 8 · 2 · 4 · 2) / (13 · 11 · 3 · 7)
= (8 · 16 · 8) / (143 · 21)
= 1024 / 3003

Sol Here, n=6 is even.
∫₀^π/2 sinⁿx dx = (n-1)/n · (n-3)/(n-2) · (n-5)/(n-4) ··· 1/2 · π/2
∫₀^π/2 sin⁶x dx = (6-1)/6 · (6-3)/(6-2) · (6-5)/(6-4) · π/2
∫₀^π/2 sin⁶x dx = 5/6 · 3/4 · 1/2 · π/2 = 5π/32

Sol: Let RHS = ∫₀^a f(a-x)dx.
Let u = a-x
du/dx = -1 ⇒ du = -dx ⇒ dx = -du
Limits: x | 0 | a
        u | a | 0
RHS = -∫_a⁰ f(u)du
{·: ∫_b^a f(x)dx = -∫_a^b f(x)dx}
RHS = ∫₀^a f(u)du
RHS = ∫₀^a f(x)dx = LHS

Sol:
∫₀⁸ f(x)dx + ∫₈¹⁰ f(x)dx = ∫₀¹⁰ f(x)dx.
∫₈¹⁰ f(x)dx = ∫₀¹⁰ f(x)dx - ∫₀⁸ f(x)dx = 17 - 12 = 5

Sol: ∫₀³ 1/√x dx = ∫₀³ 1/x^1/2 dx = ∫₀³ x^-1/2 dx.
∫₀³ 1/√x dx = [x^1/2 / (1/2)]₀³ = [2√x]₀³ = 2√3 - 0 = 2√3

Sol: Let u = 2x
du/dx = 2 ⇒ 2dx = du ⇒ dx = 1/2 du
Limit: x | 0 | 2
       u | 0 | 4
∫₀² f(2x)dx = 1/2 ∫₀⁴ f(u)du = 1/2 × 10 = 5

Sol: Let x = a sinθ then, dx/dθ = a cosθ ⇒ dx = a cosθ dθ
sinθ = x/a ⇒ θ = sin^-1(x/a)
∫ √(a²-x²) dx = ∫ √(a²-a²sin²θ) · a cosθ dθ
= ∫ √(a²(1-sin²θ)) · a cosθ dθ
= ∫ √(a²cos²θ) · a cosθ dθ
= ∫ (a cosθ)(a cosθ) dθ
= ∫ a²cos²θ dθ
= a² ∫ ( (1+cos 2θ)/2 ) dθ
= (a²/2) ∫ (1+cos 2θ) dθ
= (a²/2) [ ∫ dθ + ∫ cos 2θ dθ ]
= (a²/2) [θ + (sin 2θ)/2]
{·: sin 2θ = 2 sinθ cosθ}
= (a²/2) [θ + sinθ cosθ]
= (a²/2) [sin^-1(x/a) + (x/a) √(1-sin²θ)]
= (a²/2) [sin^-1(x/a) + (x/a) √(1-x²/a²)]
∫ √(a²-x²) dx = (a²/2) [sin^-1(x/a) + (x/a)(1-x²/a²)^1/2]

Sol: ∫ (tan x) / (sec x + cos x) dx = ∫ (sin x / cos x) / (1/cos x + cos x) dx
= ∫ (sin x / cos x) · (cos x / (1+cos²x)) dx
= ∫ (sin x) / (1+cos²x) dx
Let u = cos x ⇒ du/dx = -sin x ⇒ -du = sin x dx
= -∫ du / (1+u²) = -∫ 1 / (1+u²) du.
= -tan^-1(u)
∫ (tan x) / (sec x + cos x) dx = -tan^-1(cos x)

Sol: ∫_-1² (x²-2x)dx = ∫_-1² x²dx - 2∫_-1² x dx.
= [x³/3]_-1² - 2[x²/2]_-1²
= (8/3 - (-1/3)) - 2(4/2 - 1/2)
= (8/3 + 1/3) - 2(3/2)
= 9/3 - 2 x 5/2 = 3 - 5= -2

Sol: ∫ 1/x⁴ dx = ∫ x^-4 dx = x^-4+1 / (-4+1) = x^-3 / -3 = -1 / (3x³).

Sol: I = ∫₀¹ ∫₀² ∫₀³ [x y²z] dx dy dz.
= ∫₀¹ ∫₀² [x²y²z / 2]₀³ dy dz = ∫₀¹ ∫₀² [9y²z / 2] dy dz.
= 9/2 ∫₀¹ ∫₀² [y²z] dy dz = 9/2 ∫₀¹ [y³z / 3]₀² dz.
= 9/2 ∫₀¹ [8z / 3] dz = (9/2) × (8/3) ∫₀¹ z dz
I = 12 [z²/2]₀¹ = 12(1/2) = 6

Sol: I = ∫₁² ∫₁³ x y² dx dy = ∫₁² [x²y² / 2]₁³ dy
I = ∫₁² [9y²/2 - y²/2] dy = ∫₁² [8y²/2] dy = ∫₁² 4y² dy
I = 4 ∫₁² y² dy = 4[y³/3]₁² = 4[8/3 - 1/3] = 4 × (7/3) = 28/3

Sol: I = ∫₀¹ ∫₀² ∫₀³ [x y z] dx dy dz = ∫₀¹ ∫₀² [x²y z / 2]₀³ dy dz.
I = ∫₀¹ ∫₀² [9yz / 2] dy dz = 9/2 ∫₀¹ ∫₀² [y z] dy dz
I = 9/2 ∫₀¹ [y²z / 2]₀² dz = 9/2 ∫₀¹ [4z / 2] dz
I = 9 ∫₀¹ z dz = 9 [z²/2]₀¹ = 9(1/2) = 9/2.

Sol I = ∫₁^{log 8} ∫₀^{log y} e^x+y dx dy = ∫₁^{log 8} ∫₀^{log y} e^x · e^y dx dy
I = ∫₁^{log 8} [e^x e^y]₀^{log y} dy = ∫₁^{log 8} [e^{log y} e^y - e⁰ e^y] dy
= ∫₁^{log 8} [y e^y - e^y] dy = ∫₁^{log 8} y e^y dy - ∫₁^{log 8} e^y dy
I = ∫₁^{log 8} y e^y dy - [e^y]₁^{log 8}
I = ∫₁^{log 8} y e^y dy - [e^{log 8} - e¹] = ∫₁^{log 8} y e^y dy - (8 - e)
I₁ = ∫₁^{log 8} y e^y dy {·: ∫ u dv = uv - ∫ v du}
Let u = y ⇒ du/dy = 1 ⇒ dy = du
∫ dv = ∫ e^y dy ⇒ v = e^y
I₁ = [y e^y]₁^{log 8} - ∫₁^{log 8} e^y dy = [y e^y]₁^{log 8} - [e^y]₁^{log 8}
I₁ = [(log 8)e^{log 8} - 1e¹] - [e^{log 8} - e¹] = [8 log 8 - e] - [8 - e]
I₁ = 8 log 8 - 8
I = (8 log 8 - 8) - (8 - e) = 8 log 8 - 16 + e
∫₁^{log 8} ∫₀^{log y} e^x+y dx dy = 8 log 8 - 16 + e

Sol: I = ∫₁^a ∫₂^b 1/x · 1/y dx dy = ∫₁^a (1/y) [log x]₂^b dy
I = ∫₁^a (1/y) (log b - log 2) dy = (log b - log 2) ∫₁^a (1/y) dy
I = (log b - log 2) [log y]₁^a
I = (log b - log 2) (log a - log 1) = (log a) (log b - log 2)
∫₁^a ∫₂^b (dx dy) / (xy) = (a-1)[log b - log 2]