$$\cal{}$$$$\text{Wähle }u(x)=\sin^3x\text{ und }v'(x)=\sin x.\text{ Dann ist}$$$$u'(x)=3\sin^2x\cos x\text{ und }v(x)=-\cos x\text{ und}$$$$\begin{array}{lrl}&\int\sin^4x\mathrm dx&=-\sin^3x\cos x+\int3\sin^2x\cos^2x\mathrm dx\\&&=-\sin^3x\cos x+3\int\sin^2x(1-\sin^2x)\mathrm dx\\&&=-\sin^3x\cos x+3\int\sin^2x\mathrm dx-3\int\sin^4x\mathrm dx\\\Leftrightarrow&4\int\sin^4x\mathrm dx&=-\sin^3x\cos x+\frac32(x-\sin x\cos x)\\\Leftrightarrow&\int\sin^4x\mathrm dx&=\frac38x-\frac14\sin^3x\cos x-\frac38\sin x\cos x.\end{array}$$
