Integral sinx raise power n dx

This is a request for the integral of sin(x) raised to the power of n, with respect to x. This is typically solved using integration by parts and results in a reduction formula. Step 1: Define the integral. Let Iₙ = ∫ sinⁿ(x) dx Step 2: Use integration by parts. We can rewrite sinⁿ(x) as sinⁿ⁻¹(x) · sin(x). Let u = sinⁿ⁻¹(x) and dv = sin(x) dx. Then du = (n-1)sinⁿ⁻²(x)cos(x) dx and v = -cos(x). Step 3: Apply the integration by parts formula: ∫ u dv = uv - ∫ v du. Iₙ = sinⁿ⁻¹(x) · (-cos(x)) - ∫ (-cos(x)) · (n-1)sinⁿ⁻²(x)cos(x) dx Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1) ∫ cos²(x)sinⁿ⁻²(x) dx Step 4: Use the identity cos²(x) = 1 - sin²(x). Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1) ∫ (1 - sin²(x))sinⁿ⁻²(x) dx Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1) ∫ (sinⁿ⁻²(x) - sinⁿ(x)) dx Step 5: Split the integral. Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1) ∫ sinⁿ⁻²(x) dx - (n-1) ∫ sinⁿ(x) dx Step 6: Rearrange the equation to solve for Iₙ. Notice that the last term is -(n-1)Iₙ. Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1)Iₙ₋₂ - (n-1)Iₙ Step 7: Add (n-1)Iₙ to both sides. Iₙ + (n-1)Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1)Iₙ₋₂ nIₙ = -sinⁿ⁻¹(x)cos(x) + (n-1)Iₙ₋₂ Step 8: Divide by n to get the reduction formula. Iₙ = (-1/n)sinⁿ⁻¹(x)cos(x) + ((n-1)/n)Iₙ₋₂ The reduction formula for the integral of sinⁿ(x) dx is: ∫ sinⁿ(x) dx = (-1/n)sinⁿ⁻¹(x)cos(x) + ((n-1)/n) ∫ sinⁿ⁻²(x) dx