Determine the following integrals: 1 ∫ a³ˣ sin2x dx 2 ∫ (3 / cot⁴ 2x) dx 3 ∫ sin⁵ ax cos³ ax dx 4 ∫ (1 / (3x² - 2x + 9)) dx

Step 1: Solve integral 2.1. We need to evaluate a^3x (2x) dx. This integral requires integration by parts twice. We can use the formula for e^kx (mx) dx = e^kxk^2+m^2(k (mx) - m (mx)) + C. First, rewrite a^3x as e^3x a. So, k = 3 a and m = 2. a^3x (2x) dx = e^3x a (2x) dx Applying the formula: = e^3x a(3 a)^2 + 2^2 (3 a (2x) - 2 (2x)) + C Substitute e^3x a back to a^3x: = a^3x(3 a)^2 + 4 (3 a (2x) - 2 (2x)) + C = a^3x9 ( a)^2 + 4 (3 a (2x) - 2 (2x)) + C Step 2: Solve integral 2.2. We need to evaluate (3)/(^4(2x)) dx. Use the identity (1)/( x) = x. (3)/(^4(2x)) dx = 3 ^4(2x) dx Let u = 2x, so du = 2 dx, which means dx = (1)/(2) du. = 3 ^4 u ((1)/(2)) du = (3)/(2) ^4 u du Use the reduction formula for ^n u du = ^n-1 un-1 - ^n-2 u du. For n=4: ^4 u du = (^3 u)/(3) - ^2 u du For ^2 u du, use the identity ^2 u = ^2 u - 1: ^2 u du = (^2 u - 1) du = u - u + C_1 Substitute this back: ^4 u du = (^3 u)/(3) - ( u - u) + C_1 Now substitute this back into the main integral: (3)/(2) ( (^3 u)/(3) - u + u ) + C Substitute u = 2x: = (3)/(2) ( (^3(2x))/(3) - (2x) + 2x ) + C = (1)/(2) ^3(2x) - (3)/(2) (2x) + 3x + C = (1)/(2) ^3(2x) - (3)/(2) (2x) + 3x + C Step 3: Solve integral 2.3. We need to evaluate ^5(ax) ^3(ax) dx. Since the power of (ax) is odd, let u = (ax). Then du = a (ax) dx, so (ax) dx = (1)/(a) du. Rewrite ^3(ax) as ^2(ax) (ax) = (1 - ^2(ax)) (ax). ^5(ax) ^3(ax) dx = ^5(ax) (1 - ^2(ax)) (ax) dx Substitute u = (ax): = u^5 (1 - u^2) (1)/(a) du = (1)/(a) (u^5 - u^7) du Integrate term by term: = (1)/(a) ( (u^6)/(6) - (u^8)/(8) ) + C Substitute back u = (ax): = (1)/(a) ( (^6(ax))/(6) - (^8(ax))/(8) ) + C = (^6(ax))/(6a) - (^8(ax))/(8a) + C Step 4: Solve integral 2.4. We need to evaluate (1)/(3x^2 - 2x + 9) dx. Complete the square in the denominator. 3x^2 - 2x + 9 = 3(x^2 - (2)/(3)x +