Mathematics

Solve :
$$I=\int \sin^{6}{x}dx$$


SOLUTION
$$\displaystyle I = \int sin^{6}x\,dx $$ 
$$\displaystyle = \int (sin^{2}x)^{2}(sin^{2}x)dx $$
$$\displaystyle = \int (\frac{1-cos2x}{2})^{2}(\frac{1-cos2x}{2})dx $$
$$\displaystyle = \frac{1}{8}\int (1-2cos2x+cos^{2}x)(1-cos2x)dx $$
$$\displaystyle = \frac{1}{8}\int (1-2cos2x+\frac{1+cos4x}{2})(1-cos2x)dx $$
$$\displaystyle = \frac{1}{16}\int (2-4cos2x+cos4x)(1-cos2x)dx $$
$$\displaystyle = \frac{1}{16}\int (3-4cos2x+cos4x)(1-cos2x)dx $$
$$\displaystyle = \frac{1}{16}\int (3-4cos2x-3cos2x+cos4x+4cos^{2}2x-cos4xcos2x)dx $$
$$\displaystyle = \frac{1}{16}\int [3-7cos2x+cos4x+2\times 2cos^{2}2x-\frac{1}{2}.2cos^{4}xcos2x]dx $$
$$\displaystyle = \frac{1}{16}[3-7cos2x+cos^{4}x+2(1+cos^{4}x)-\frac{1}{2}(cos6x+cos2x)]dx $$
$$\displaystyle = \frac{1}{16}(3-7cos2x+cos^{4}x+2+2cos^{4}x-\frac{1}{2}(cos6x+cos2x))dx $$
$$\displaystyle = \frac{1}{32}\int [6-14cos2x+2cos4x+4+4cos4x-cos6x-cos2x]dx $$
$$\displaystyle = \frac{1}{32}[10-15cos2x+6cos4x-cos6x]dx $$
$$\displaystyle \frac{1}{32}[10-15cos2x+6cos4x-cos6x]dx $$ 
$$\displaystyle \frac{1}{32}[10x-\frac{15sin2x}{2}+\frac{6sin4x}{4}-\frac{sin6x}{6}]dx $$
$$\displaystyle \frac{1}{32}[10x-\frac{15sin2x}{2}+\frac{3}{2}sin4x] $$
$$\displaystyle \frac{1}{92}[60x-45sin2x+9sin4x-sin6x]+c $$ 
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Subjective Medium Published on 17th 09, 2020
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