Question

Evaluate the following integrals using Integration by Parts (IBP). Show clear support for you answer. a. ∫sin ⁻¹xdx b. ∫3x(lnx) ²dx c. ∫e ⁴ˣcos(2x)dx

Answer 1

The integral ∫sin⁻¹xdx is equal to xsin⁻¹x + √(1-x²) + C, where C is the constant of integration. The correct option is a. ∫sin⁻¹xdx

Step-by-step explanation:

To evaluate the integral ∫sin⁻¹xdx, we use integration by parts (IBP). Let u = sin⁻¹x and dv = dx. Then, du = 1/√(1-x²)dx and v = x. Applying the IBP formula ∫udv = uv - ∫vdu, we get:

∫sin⁻¹xdx = xsin⁻¹x - ∫x(1/√(1-x²))dx

To solve the remaining integral, we can make a substitution. Let w = 1 - x², then dw/dx = -2x, and dx = -dw/(2√w). Substituting these into the integral, we get:

-∫x(1/√(1-x²))dx = -∫x(-dw/(2√w)) = 1/2 ∫dw/√w

Integrating w⁻¹/₂ with respect to w gives 2√w. Substituting back in terms of x, we get 2√(1-x²). Therefore, the final result is:

∫sin⁻¹xdx = xsin⁻¹x + 2√(1-x²) + C

This is the antiderivative, and we add the constant of integration, denoted as C, to account for the family of functions that differ by a constant. The correct option is a.

Answer 2

To evaluate the given integrals using Integration by Parts (IBP), we can apply the formula ∫u dv = uv - ∫v du. Using this formula, we can evaluate each integral step-by-step.

Step-by-step explanation:

To evaluate the given integrals using Integration by Parts (IBP), we need to use the formula:

∫u dv = uv - ∫v du

a. For the integral ∫sin-1(x) dx, we can let u = sin-1(x) and dv = dx.

This gives du = 1/√(1-x2) dx and v = x. Applying the formula, we have:

∫sin-1(x) dx = x*sin-1(x) - ∫x / √(1-x2) dx

b. For the integral ∫3x(lnx)2 dx, we can let u = (lnx)2 and dv = 3x dx.

This gives du = (2/lnx)(lnx) dx and v = (3/2)x2.

Applying the formula, we have:

∫3x(lnx)2 dx = (3/2)x2(lnx)2 - ∫(3/2)x2(2/lnx)(lnx) dx

c. For the integral ∫e4xcos(2x) dx, we can let u = e4x and dv = cos(2x) dx.

This gives du = 4e4x dx and v = (1/2)sin(2x).

Applying the formula, we have:

∫e4xcos(2x) dx = (1/2)e4xsin(2x) - ∫(1/2)(4e4x)sin(2x) dx