Question

Integral: 1) arccos^3 xdx/(1-x^2)^(1/2) 2) sin4x(cos4x)^5 dx 3) (6x-5)sin2xdx

Answer 1

Решение на фотографии

Answer 2

Sure, let's solve each integral step by step:

1. ∫(arccos^3 x)dx / (1-x^2)^(1/2)

To solve this integral, we can perform a u-substitution. Let's let u = arccos(x). Then, du = -1/√(1-x^2) dx.

Now, the integral becomes:

∫(u^3)(-du) = -∫u^3 du = -u^4/4 + C,

where C is the constant of integration.

Finally, replace u with arccos(x):

∫(arccos^3 x)dx / (1-x^2)^(1/2) = -arccos^4 x / 4 + C.

2. ∫sin(4x)(cos(4x))^5 dx

To solve this integral, we can perform a u-substitution. Let's let u = cos(4x). Then, du = -4sin(4x) dx.

Now, the integral becomes:

∫-u^5 / 4 du = -∫u^5 / 4 du = -u^6 / 24 + C,

where C is the constant of integration.

Finally, replace u with cos(4x):

∫sin(4x)(cos(4x))^5 dx = -cos^6(4x) / 24 + C.

3. ∫(6x-5)sin(2x) dx

To solve this integral, we can use integration by parts. Integration by parts states:

∫u dv = uv - ∫v du.

Let's choose u = (6x - 5) and dv = sin(2x) dx. Then, du = 6 dx and v = -1/2 cos(2x).

Now, apply the formula:

∫(6x-5)sin(2x) dx = -1/2 (6x - 5) cos(2x) - ∫(-1/2 cos(2x) * 6) dx = -1/2 (6x - 5) cos(2x) + 3∫cos(2x) dx = -1/2 (6x - 5) cos(2x) + 3 * (1/2) * sin(2x) + C = -3x cos(2x) + (5/2) cos(2x) + (3/2) sin(2x) + C,

where C is the constant of integration.

So the final result is ∫(6x-5)sin(2x) dx = -3x cos(2x) + (5/2) cos(2x) + (3/2) sin(2x) + C.