sanuluy

2021-10-13

Evaluate the following integrals.
$\int {\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}xdx$

Faiza Fuller

We have to evaluate the integral:
$\int {\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}xdx$
Since we know the identity, $2\mathrm{sin}x\mathrm{cos}x=\mathrm{sin}2x$
Therefore multiplying and dividing by ${2}^{2}$,
$\frac{{2}^{2}}{{2}^{2}}\int {\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}xdx=\frac{1}{4}\int {2}^{2}{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}xdx$
$=\frac{1}{4}\int {\left(2\mathrm{sin}x\mathrm{cos}x\right)}^{2}dx$
$=\frac{1}{4}\int {\left(\mathrm{sin}2x\right)}^{2}dx$
$=\frac{1}{4}\int {\mathrm{sin}}^{2}2xdx$
We have identity,
${\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x=\mathrm{cos}2x$
$\mathrm{cos}2x=1-2{\mathrm{sin}}^{2}x\left({\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x=1\right)$
${\mathrm{sin}}^{2}x=\frac{1-\mathrm{cos}2x}{2}$
Therefore replacing x to 2x, we get
${\mathrm{sin}}^{2}2x=\frac{1-\mathrm{cos}4x}{2}dx$
$=\frac{1}{8}\int \left(1-\mathrm{cos}4x\right)dx$
$=\frac{1}{8}\left(\int dx-\int \mathrm{cos}4xdx\right)$
$=\frac{1}{8}x-\frac{1}{8}\left(\frac{\mathrm{sin}4x}{4}\right)+C$
$=\frac{1}{8}\left(x-\frac{\mathrm{sin}4x}{4}\right)+C$
Where, C is an arbitrary constant.
Hence, value of integral is $\frac{1}{8}\left(x-\frac{\mathrm{sin}4x}{4}\right)+C$
Note: we have used following formula,
$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$
$\int \mathrm{cos}xdx=\mathrm{sin}x+C$

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