Question

integration sin square x cos square x sinx dx​

Answer 1

Answer:

Step-by-step explanation:

$\int\limits {sin^2x\;cos^2x\;sinx} \, dx\\ \\\int\limits {sin^3x(1-sin^2x)} \,dx\\\\\int\limits{sin^3x}\, dx\;-\int{sin^5x}\,dx\\\\WKT\\\\\int{sin^nx}\,dx=-\frac{1}{n}cosx.x.sin^{n-1}x+\frac{n-1}{n}\int{sin^{n-2}x}\,dx\\\\\int{sin^3x}\,dx-(-\frac{1}{5}cosx.x.sin^{4}x+\frac{4}{5}\int{sin^{3}x}\,dx)\\\\\frac{1}{5}\int{sin^3x}\,dx+\frac{x}{5}cosx.sin^4x\\\\\frac{1}{5}(-\frac{1}{3}cosx.x.sin^{2}x+\frac{2}{3}\int{sin^{}x}\,dx)+\frac{x}{5}cosx.sin^4x$

$\frac{x}{5}cosx.sin^4x-\frac{x}{15}cosx.sin^2x-\frac{2}{15}cosx$

Hope it Helps