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Question Number 162804 by mnjuly1970 last updated on 01/Jan/22

$$$$$$$$$$\mathrm{\Omega }=\int {sin}^2\left(x\right).{cos}^4\left(x\right)dx$$$$$$

Answered by mindispower last updated on 01/Jan/22

$${sin}^2\left(x\right){cos}^4\left(x\right)=\frac{1}{4}{sin}^2\left(2x\right){cos}^2\left(x\right)$$$$=\frac{1}{4}\left(\frac{1-cos\left(4x\right)}{2}\right)\left(\frac{1+cos\left(4x\right)}{2}\right))$$$$=\frac{1}{16}\left({sin}^2\left(4x\right)\right)=\frac{1-cos\left(8x\right)}{32}$$$$\int \frac{1-cos\left(8x\right)}{32}dx=\frac{x}{32}-\frac{sin\left(8x\right)}{256}+c$$

Commented by tounghoungko last updated on 02/Jan/22

$$why\cos {}^{2}x\stackrel{?}{=}\frac{1+\cos 4x}{2}$$

Answered by john_santu last updated on 01/Jan/22

$$\mathrm{\Omega }=\frac{1}{4}\int \sin {}^{2}2x\cos {}^{2}xdx$$$$\mathrm{\Omega }=\frac{1}{16}\int {\left(2\sin 2x\cos x\right)}^{2}dx$$$$\mathrm{\Omega }=\frac{1}{16}\int {(\sin 3x+\sin x)}^{2}dx$$$$\mathrm{\Omega }=\frac{1}{16}\int [\frac{1-\cos 6x+1-\cos 2x}{2}+2\sin 3x\sin x]dx$$$$\mathrm{\Omega }=\frac{1}{32}(2x-\frac{1}{6}\sin 6x-\frac{1}{2}\sin 2x)-\frac{1}{16}\int (\cos 4x-\cos 2x)dx$$$$\mathrm{\Omega }=\frac{x}{16}-\frac{\sin 6x}{192}-\frac{\sin 2x}{64}-\frac{\sin 4x}{64}+\frac{\sin 2x}{32}+c$$$$\mathrm{\Omega }=\frac{x}{16}-\frac{\sin 6x}{192}+\frac{\sin 2x}{64}-\frac{\sin 4x}{64}+c$$

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