Esercitazione guidata Analisi Matematica 1 - nona parte

Integrali

1)∫ ^{senx cosx}/_{sen⁴x + cos⁴x} dx= ∫ ^{senx cosx}/_{cos³x (sen⁴x/cos⁴x + 1)} dx= ∫ ^senx/_{cos³x (tg⁴x + 1)} dx= ∫ ^senx/_cosx · 1/cos²x · 1/(tg⁴x + 1) dx= ∫ ^tgx/_cos²x · 1/(tg⁴x + 1) dx

∫ ^f'(x)/_{1 + f²(x)} dx = arctg f(x) + C= 1/2 ∫ (2 tg x/cos²x) · 1/¹/_{1 + (tg²x)²} dx= 1/2 arctg (tg²x) + C

2)∫ ^{3x + 1}/_{x² - 6x + 5} dxx² - 6x + 5 = (x - 5)(x - 1)= ∫ ^{3x + 1}/_{x² - 6x + 5} = ∫ A/(x - 5) + ∫ B/(x - 1)= ^{A(x - 1) + B(x - 5)}/_{(x - 5)(x - 1)}= ^{Ax - A + Bx - 5B}/_{(x - 5)(x - 1)}= ^{x(A + B) + (-A - 5B)}/_{(x - 5)(x - 1)}

1)∫ ^{sen x cos x}/_{sen⁴ x + cos⁴ x} dx= ∫ ^{sen x cos x}/_{cos³ x} ( ^{sen⁴ x}/_{cos⁴ x} + 1 ) dx= ∫ ^{sen x}/_{cos³ x (tg⁴ x + 1)} dx= ∫ ^{sen x}/_{cos x} . 1/_{cos² x} . 1/_{tg⁴ x + 1} dx= ∫ ^{tg x}/_{cos² x} . 1/_{tg⁴ x + 1} dx

∫ ^{f '(x)}/_{1 + f²(x)} dx = arctg f(x) + C= 1/2 ∫ ^{2 tg x}/_{cos² x} . 1/_{1 + (tg² x)²} dx= 1/2 arctg (tg² x) + C

2)∫ ^{3 x + 1}/_{x² - 6 x + 5} dxx² - 6 x + 5 = (x - 5)(x - 1)= ∫ ^{3 x + 1}/_{x² - 6 x + 5} = ∫ A/_{x - 5} + ∫ B /_{x - 1} A, B ∈ ℝ= ^{A(x - 1) + B(x - 5)}/_{(x - 5)(x - 1)}= ^{x(A + B) + (-A - 5B)}/_{(x - 5)(x - 1)}

A + B = 3-A - 5B = 1A = 4B = -1

∫ ^{3x + 1}⁄_{x² - 6x + 5} dx = ∫ ⁴⁄_{x - 5} dx - ∫ ¹⁄_{x - 1} dx= 4 log|x - 5| - log|x - 1| + C

∫ ^f'(x)⁄_f(x) dx = log|f(x)| + C

3) Integrale con frazioni parziali

∫ ^{3x + 2}⁄_{x(x² + 1)} dx         x(x² + 1) = 0         A, B, C ∈ ℝ              X = 0         V         x² + 1 = 0

^{3x + 2}⁄_{x(x² + 1)} = ^A⁄_x + ^{Bx + C}⁄_{x² + 1} = ^{A(x2 + 1) + (Bx + C)x}⁄_{x(x² + 1)}= ^{Ax2 + A + Bx2 + Cx}⁄_{x(x² + 1)} = ^{x2(A + B) + Cx + A}⁄_{x(x² + 1)}

A + B = 0C = 3A = 2B = -2C = 3A = 2

∫ ^{3x + 2}⁄_{x(x² + 1)} dx = ∫ ²⁄_x dx + ∫ ^{-2x + 3}⁄_{x² + 1} dx= 2 ∫ ¹⁄_x dx - ∫ ^2x⁄_{x² + 1} dx + 3 ∫ ¹⁄_{x² + 1} dx= 2 log|x| - log(x² + 1) + 3 arctg x + C

4) Integrazione per parti

∫ x²cos x dx

∫ f'(x)g(x) dx = f(x)g(x) - ∫ f(x)g'(x) dxf'(x) = cos xg(x) = x²f(x) = sen xg'(x) = 2x

∫ x²cos x dx = x²sen x - ∫ 2x sen x dx= x²sen x - 2 ∫ x sen x dxf' = sen xg = xf = -cos xg' = 1= x²sen x - 2 { -x cos x - ∫ -cos x dx }= x²sen x + 2 x cos x - 2 ∫ cos x dx= x²sen x + 2 x cos x - 2 sen x + C= (x² - 2) sen x + 2x cos x + C

5) Integrale definito

∫₀^√3 4 |x-1| arctg x dx

|x-1| = { x - 1 \ se \ x ≥ 1-(x-1) \ se \ x < 1 = ∫₀¹ 4(-(x-1)) arctg x dx + ∫₁^√3 4(x-1) arctg x dx = -4 ∫₀¹ (x-1) arctg x dx + 4 ∫₁^√3 (x-1) arctg x dx

∫(x-1)arctgx dx= (x²/2 - x) arctgx - ∫(x²/2 - x) 1/(1+x²) dx= (x²/2 - x) arctgx - 1/2 ∫{ x²-2x/x²+1 } dx= (x²/2 - x) arctgx - 1/2 ∫{ x²/x²+1 dx - ∫{ 2x/x²+1 dx }= (x²/2 - x) arctgx - 1/2 ∫{ x²+1-1/x²+1 } dx - ∫{ 2x/x²+1 dx }= (x²/2 - x) arctgx - 1/2 ( ∫ x²+1/x²+1 dx - ∫ 1/x²+1 dx )= (x²/2 - x) arctgx - 1/2 x + 1/2 arctgx + 1/2 log(x²+1) + C= (x²/2 - x + 1/2) arctgx - 1/2 x + 1/2 log(x²+1) + C

∫₀¹ (x-1) arctgx dx= [(x²/2 - x + 1/2) arctgx - 1/2 x + 1/2 log(x²+1)] ₀¹= (0 - 1/2 + 1/2 log 2) - (0) = - 1/2 + 1/2 log 2

∫₁^√3 (x-1) arctg x dx= [ (x²/2 - x + 1/2) arctg x - 1/2 x + 1/2 log (x² + 1) ]₁^√3= [ (2-√3)π/3 - √3/2 + 1/2 log 4 ] - ( -1/2 + 1/2 log 2 )= (2-√3)π/3 - √3/2 + 1/2 log 2 + 1/21/2 log 4 = 1/2 log (2²) = 1/2 log 2² = log 2

∫₁^√3 4x-1 arctg x dx == -4 ∫₁^√3 (x-1) arctg x dx + 4 ∫₁^√3 (x-1) arctg x dx= -4 [ -1/2 + 1/2 log 2 ) + 4 [ (2-√3)π/3 - √3/2 + 1/2 log 2 + 1/2 ]= 2 - 2 log 2 + 4 (2-√3)π/3 - 2√3 + 2 log 2 + 2= 4 + 4 (2-√3)π/3 - 2√3

6) Sostituzione

6) ∫₉¹⁶ (√x - 3)/(x - 3√x + 2) dx = ∫₃⁴ (t - 3)/(t² - 3t + 2) ⋅ 2t dt

√x = tx = (t²)dx = 2t dt

∫₂⁶ f(x) dx = ∫_φ^-1(a)^φ-1(b) f(φ(t)) φ'(t) dt

∫₃⁴ (2t² - 6t) / (t² - 3t + 2) dt= ∫₃⁴ (2 (t² - 3t)) / (t² - 3t + 2) dt= 2 ∫₃⁴ (t² - 3t + 2 - 2) / (t² - 3t + 2) dt= 2 ∫₃⁴ 1 dt - 4 ∫₃⁴ 1 / (t² - 3t + 2) dt= 2 [t]₃⁴ - 4 ∫₃⁴ 1 / ((t - 2)(t - 1)) dt= 2 [4 - 3] - 4 ∫₃⁴ 1 / ((t - 2)(t - 1)) dt

I = 2 + 4 log 3 - 8 log 2

Integrali successivi

a) ∫₀^π/4 1 / (3tgx + 2cosx) dx= ∫₀^π/4 1 / (3senx / cosx + 2cosx) dx= ∫₀^π/4 cosx / (3senx + 2cos²x) dx

senx = tcosx dx = dt= ∫₀^√2/2 dt / (3t + 2(1 - t²))= ∫₀^√2/2 -1 / (2t² + 3t + 2) dt= -∫_√2/2⁰ 1 / (2t² - 3t - 2) dt= 1/5 [log(√2 + 1) - log(1/2)] - 1/5 [log(√2/2) - 2 (-log 2)]

8) ∫₃⁴ eˣ / (e²ˣ - 3eˣ + 2) dxeˣ = t     x = log t     dx = 1/t dt= ∫_e³^e⁴ t / (t² - 3t + 2) . 1/t dt= ∫_e³^e⁴ 1 / (t² - 3t + 2) dt

9) ∫₀^π/4 2 / (1 + tg x)² dx