1. **Problem Statement:** We need to find the reduction formula for the integral $$I_m = \int \cos^m x \, dx$$ and show that $$I_{m,n} = -\frac{\cos^{n-1} x \sin x}{m+n} + \frac{m}{m+n} I_{m-1,n-1}$$ Also, solve the integral $$I = \int \sin^2 x \, dx$$. 2. **Step 1: Deriving the reduction formula for $$I_{m,n} = \int \cos^m x \sin^n x \, dx$$** - Use integration by parts. Let $$u = \cos^m x, \quad dv = \sin^n x \, dx$$ - Differentiate and integrate: $$du = -m \cos^{m-1} x \sin x \, dx, \quad v = -\frac{\cos^{n-1} x}{n}$$ (using substitution for $v$) - Applying integration by parts: $$I_{m,n} = uv - \int v \, du = -\frac{\cos^{n-1} x \sin x}{m+n} + \frac{m}{m+n} \int \cos^{m-1} x \sin^{n-1} x \, dx$$ - This gives the reduction formula: $$I_{m,n} = -\frac{\cos^{n-1} x \sin x}{m+n} + \frac{m}{m+n} I_{m-1,n-1}$$ 3. **Step 2: Solving $$I = \int \sin^2 x \, dx$$** - Use the identity: $$\sin^2 x = \frac{1 - \cos 2x}{2}$$ - Substitute into the integral: $$I = \int \sin^2 x \, dx = \int \frac{1 - \cos 2x}{2} \, dx = \frac{1}{2} \int (1 - \cos 2x) \, dx$$ - Integrate term by term: $$I = \frac{1}{2} \left( x - \frac{\sin 2x}{2} \right) + C = \frac{x}{2} - \frac{\sin 2x}{4} + C$$ **Final answers:** - Reduction formula: $$I_{m,n} = -\frac{\cos^{n-1} x \sin x}{m+n} + \frac{m}{m+n} I_{m-1,n-1}$$ - Integral of $$\sin^2 x$$: $$\int \sin^2 x \, dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$$