# Solve this.  Solve this.

dear student

$I=\int {\mathrm{sin}}^{2}x{\mathrm{cos}}^{4}xdx\phantom{\rule{0ex}{0ex}}=\int \frac{1-\mathrm{cos}2x}{2}{\left(\frac{1+\mathrm{cos}2x}{2}\right)}^{2}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{8}\int \left(1-\mathrm{cos}2x\right)\left(1+{\mathrm{cos}}^{2}2x+2\mathrm{cos}2x\right)dx\phantom{\rule{0ex}{0ex}}=\frac{1}{8}\int \left(1+{\mathrm{cos}}^{2}2x+2\mathrm{cos}2x-\mathrm{cos}2x-{\mathrm{cos}}^{3}2x-2{\mathrm{cos}}^{2}2x\right)dx\phantom{\rule{0ex}{0ex}}=\frac{1}{8}\int \left(1-{\mathrm{cos}}^{2}2x+\mathrm{cos}2x-{\mathrm{cos}}^{3}2x\right)dx\phantom{\rule{0ex}{0ex}}=\frac{1}{8}\int \left(1-\frac{1+\mathrm{cos}4x}{2}+\mathrm{cos}2x-{\mathrm{cos}}^{2}2x.\mathrm{cos}2x\right)dx\phantom{\rule{0ex}{0ex}}=\frac{1}{8}\left(x/2-\mathrm{sin}4x/8+\mathrm{sin}2x/2-\int \left(1-{\mathrm{sin}}^{2}2x\right)\mathrm{cos}2xdx\right)+c\phantom{\rule{0ex}{0ex}}=\frac{1}{8}\left(x/2-\mathrm{sin}4x/8+\mathrm{sin}2x/2-\frac{1}{2}\left(\mathrm{sin}2x-\frac{{\mathrm{sin}}^{3}2x}{3}\right)\right)+c$

regards

• 0
What are you looking for?