Question

∫dx/√(7-5x^2 ) ∫(1-3x)×cos⁡5xdx=-3∫x×cos⁡(5x)dx ∫(1-2x)×e^3x dx

Answer 1

Ответ:

Пошаговое объяснение:

Answer 2

Let's solve each integral step by step:

1. ∫ dx/√(7-5x^2)

This integral can be solved using a trigonometric substitution. Let's use x = √(7/5)sinθ. Then, dx = √(7/5)cosθdθ, and we can rewrite the integral as follows:

∫ dx/√(7-5x^2) = ∫ √(7/5)cosθdθ / √(7-5(√(7/5)sinθ)^2)

Simplify the expression inside the square root:

√(7-5(√(7/5)sinθ)^2) = √(7-5(7/5)sin^2θ) = √(7-7sin^2θ) = √(7(1-sin^2θ)) = √(7cos^2θ)

Now the integral becomes:

∫ dx/√(7-5x^2) = ∫ √(7/5)cosθdθ / √(7cos^2θ)

Cancel out the common terms:

∫ dx/√(7-5x^2) = ∫ dθ / √5

Integrate with respect to θ:

∫ dx/√(7-5x^2) = (1/√5)θ + C

Substitute back θ = arcsin(x/√(7/5)):

∫ dx/√(7-5x^2) = (1/√5)arcsin(x/√(7/5)) + C1

2. ∫ (1-3x)×cos⁡5x dx

This integral can be solved using integration by parts. Let u = 1-3x and dv = cos(5x)dx, then du = -3dx and v = (1/5)sin(5x). The formula for integration by parts is ∫u dv = uv - ∫v du.

∫ (1-3x)×cos⁡5x dx = (1-3x)×(1/5)sin(5x) - ∫ (1/5)sin(5x)×(-3) dx

∫ (1-3x)×cos⁡5x dx = (1/5)(1-3x)sin(5x) + (3/5)∫sin(5x) dx

Integrate the remaining integral:

∫ (1-3x)×cos⁡5x dx = (1/5)(1-3x)sin(5x) - (3/5)(-1/5)cos(5x) + C2

Simplify:

∫ (1-3x)×cos⁡5x dx = (1/5)(1-3x)sin(5x) + (3/25)cos(5x) + C2

3. ∫ (1-2x)×e^3x dx

This integral can be solved using integration by parts as well. Let u = (1-2x) and dv = e^(3x)dx, then du = -2dx and v = (1/3)e^(3x).

∫ (1-2x)×e^3x dx = (1-2x)(1/3)e^(3x) - ∫ (1/3)e^(3x)×(-2)dx

∫ (1-2x)×e^3x dx = (1/3)(1-2x)e^(3x) + (2/3)∫e^(3x)dx

Integrate the remaining integral:

∫ (1-2x)×e^3x dx = (1/3)(1-2x)e^(3x) + (2/9)e^(3x) + C3

Simplify:

∫ (1-2x)×e^3x dx = (1/3)(1-2x)e^(3x) + (2/9)e^(3x) + C3

Now we have solved all three integrals.