Answer
$$\int {{{\sin }^n}xdx} = - \frac{1}{n}{\sin ^{n - 1}}x\cos x + \frac{{n - 1}}{n}\int {{{\sin }^{n - 2}}xdx} $$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^n}x} dx \cr
& {\text{Rewrite the integrand}} \cr
& \int {{{\sin }^n}x} dx = \int {{{\sin }^{n - 1}}x \cdot \sin x} dx \cr
& {\text{Let }}u = {\sin ^{n - 1}}x{\text{ }} \Rightarrow {\text{ }}du = \left( {n - 1} \right){\sin ^{n - 2}}x\cos xdx \cr
& dv = \sin xdx{\text{ }} \Rightarrow {\text{ }}v = - \cos x \cr
& {\text{then, }} \cr
& \int {{{\sin }^n}x} dx = - \cos x \cdot {\sin ^{n - 1}}x - \int {\left( { - \cos x} \right)\left( {n - 1} \right){{\sin }^{n - 2}}x\cos xdx} \cr
& \int {{{\sin }^n}x} dx = - \cos x{\sin ^{n - 1}}x + \left( {n - 1} \right)\int {{{\cos }^2}x{{\sin }^{n - 2}}xdx} \cr
& {\text{Simplifying}} \cr
& \int {{{\sin }^n}x} dx = - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right)\int {\left( {1 - {{\sin }^2}x} \right){{\sin }^{n - 2}}xdx} \cr
& \int {{{\sin }^n}x} dx = - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right)\int {{{\sin }^{n - 2}}xdx - } \left( {n - 1} \right)\int {{{\sin }^n}xdx} \cr
& \int {{{\sin }^n}x} dx + \left( {n - 1} \right)\int {{{\sin }^n}xdx} = - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right)\int {{{\sin }^{n - 2}}xdx} \cr
& {\text{Factoring}} \cr
& \left[ {1 + n - 1} \right]\int {{{\sin }^n}xdx} = - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right)\int {{{\sin }^{n - 2}}xdx} \cr
& n\int {{{\sin }^n}xdx} = - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right)\int {{{\sin }^{n - 2}}xdx} \cr
& {\text{Solving for }}\int {{{\sin }^n}xdx} \cr
& \int {{{\sin }^n}xdx} = - \frac{1}{n}{\sin ^{n - 1}}x\cos x + \frac{{n - 1}}{n}\int {{{\sin }^{n - 2}}xdx} \cr} $$
