prove this $\displaystyle \int{sin^3xcos^nx}d(x)=\frac{2}{3+n}\int sinxcos^nxd(x)-\frac{sin^2xcos^{n+1}x}{3+n}$ In this problem i write $\displaystyle sin^3x$ like this $\displaystyle sin^3x=\frac{3sinx-sin3x}{4}$after that i use $\displaystyle \int u \frac{d(v)}{d(x)}$ format and after that i get stucked and i get everything in cos.please give me a solution??