Prove d'esame: Integrali e Limiti svolti

1) _x ∫ arcsen ^k_y ⁿ_x dx

∫ arcsen ^k_y dx Pongo t = k_x

∫ arcsen t dt Pongo g = arcsen t + t_x

∫ x²cos y dy = x²sen y + x²y₂ cos y dy

∫ x²cos y dy = x²sen y - ∫ y²cos y dy

Quindi

∫ x²cos y dy = x²sen y + x ^x2 - ∫ x²cos y dy

∫ ( ∫ ¿ ) ² = t^{(cos y + sen y)} / 2

Quindi

∫ _x arcsen ^k_y dx = l arcsen^k_yn (cos(arcsen ^k_y) + sen (arcsen y)) / 2

+ c

1) ∫ _x avcsen ^x₂ dx

∫ avcsen ^x₂ dx = Posg t_o = avcsen² t

∫ avcsen dt

Posg g = avcsen² + to_t

∫ x²cos y dy = x²sen y + ∫ y²sen y dy

∫ y²sen y dy = x²sen y - ∫ y²cos y dy

Quindi

∫ x²cos y dy = x²sen y + x²y² - ∫ x²cos y dy

∫ x ²cos y dy = x² (cos y + sen y) / 2

Quindi

∫ avcsen ^t_t dx = e avcsen ^x₂ (cos(avcsen k) + sen(avcsen ^x₂)) / 2

+C

1. ∫ x³ + x² x⁴ / (x³ + x²) dx = ∫ x x⁴ dx - ∫ x⁴ x³ + ∫ x³ dx + ∫ x³ dx

   = x⁴ x/2 + ∫ ^x x² x_3/2 ∫ x dx

   = ∫ x x x dx + ∫ x x ∫ x x ∫ x dx + ∫ ¹ / 1 + x ⁻ dx

   = ∫ x x∫dx + x² x² x∫dx - x∫ x dx + ∫ x x x dx + c

2. ∫ 2x dx = ∫ x x x ∫ x dx = x² x /2 + ∫ x dx + c.

∫ x lg (√x+1 + √x-1) dx

∫ x lg (√x+1 + √x-1) dx = x²/2 lg (√x+1 + √x-1) - x/2 ∫ 1/(√x+1 - √x-1) 4/2√x+1 - √x-1

∫ x lg (√x+1 + √x-1) dx = x²/2 lg (√x+1 + √x-1) - x/2 ∫ 1/(√x+1 + √x-1)

Sia che ho derivato D (√x-1 - √x+1) = (1/√x-1 - 1/√x+1) =

= 2√x+1 √x-1 - x/√x+1 √x-1 - x/√x+1 √x-1

Pertanto

∫ x/√x+√x+1 dx = x√x - √x+1 - √x-1 dx

Sostituire

√x-1 √x+1 x : ∫ y/√y-1 dy : ∫ y/√y-1 dy = ∫ y²/√y-1 dy = ∫ y²/√y-1 y

Pertanto

∫ lg (√x+1, + √x-1) dx = x²/2 lg (√x+1, + √x-1) -1/4 x√x-1 √x+1 - √x+1 - arcsin √x+1 + C

1) ∫ arcsen √x dx

Pong √x = y , x = y⁴ dx = y³dy

∫₀¹ arcsen y⁴ y³ dy = y⁴ arcsen y - ∫₀¹ √⁴_1-y²

∫₀¹ y √⁴_1-y² dy = ∫₀¹ y dq √⁴_1-y² dy

S. che D (√^-1_1-y²) = ¹ _{2 √^-1 1-y²} (-dy) Pertanto,

∫₀¹ y √⁴_1-y² dy = √ √^-1_1-y2 , ∫ ³₁ √⁴_1-y2 dy

Pong √^-1_1-y² = z , y = 1-z² , y = √ √^-1_z² dy = ^{- 2z}_{2 √^-12}

Quindi:

∫₀¹ y √⁴_1-y² dy =³₀(x+z)^1/2 dz = ∫₀¹ √√ _2z² dz =^{- 2√1_1-z2}, d(1-z²) =

=∫₃²(x+z)^1/2 =₂¹ = ³₅(z-z^23/2)

Riscrivendo

∫₀¹ arcsen √x dx =[x arcsen √₀] √ ^{2 0}₁ , (√^-1₁ √^-1₁ (1- √³_√0 )^-1)

a) \(\lim_{x \to 0} \frac{x^2 \cdot \lg^2(4+\sqrt{x}) - x}{\sqrt{\cos 3x}} = 0 \)

\[\lg^2 (4+\sqrt{x}) = \lg (4+\sqrt{x}) \cdot \lg (4+\sqrt{x})\]

\(\lim_{x \to 0} \frac{x^2 \cdot \lg^2(4+\sqrt{x}) - x}{\sqrt{\cos 3x}} = \lim_{x \to 0} \frac{x^2 \cdot \lg (4+\sqrt{x}) \cdot \lg (4+\sqrt{x}) - x}{\sqrt{\cos 3x}}\)

\(= \lim_{x \to 0} \left( \frac{x^2}{x} \cdot \frac{\lg (4+\sqrt{x})}{\sqrt{x}} \cdot \lg (4+\sqrt{x}) \right) =\)

\(\lim_{x \to 0} \frac{x}{\sqrt{\cos 3x}} \cdot \lim_{x \to 0} \frac{x^2 + 1}{\sqrt{x(4 - \cos x)}} = \lim_{x \to 0} \frac{x}{\sqrt{4 - \cos x}} \)

\(= \lim_{x \to 0} \frac{\sqrt{x}}{\sqrt{4 - \cos x}} \cdot \lim_{x \to 0} \frac{\sqrt{x}}{\sqrt{4 - \cos x} - x} = \lim_{x \to 0} \frac{\sqrt{x}}{\sqrt{4/2}} = 0\)