sanuluy

2021-10-13

Evaluate the following integrals.

$\int {\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}xdx$

Faiza Fuller

Skilled2021-10-14Added 108 answers

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,

${\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x=\mathrm{cos}2x$

$\mathrm{cos}2x=1-2{\mathrm{sin}}^{2}x({\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x=1)$

$\mathrm{sin}}^{2}x=\frac{1-\mathrm{cos}2x}{2$

Therefore replacing x to 2x, we get

${\mathrm{sin}}^{2}2x=\frac{1-\mathrm{cos}4x}{2}dx$

$=\frac{1}{8}\int (1-\mathrm{cos}4x)dx$

$=\frac{1}{8}(\int dx-\int \mathrm{cos}4xdx)$

$=\frac{1}{8}x-\frac{1}{8}\left(\frac{\mathrm{sin}4x}{4}\right)+C$

$=\frac{1}{8}(x-\frac{\mathrm{sin}4x}{4})+C$

Where, C is an arbitrary constant.

Hence, value of integral is$\frac{1}{8}(x-\frac{\mathrm{sin}4x}{4})+C$

Note: we have used following formula,

$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$

$\int \mathrm{cos}xdx=\mathrm{sin}x+C$

Since we know the identity,

Therefore multiplying and dividing by

We have identity,

Therefore replacing x to 2x, we get

Where, C is an arbitrary constant.

Hence, value of integral is

Note: we have used following formula,

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