The Notes: `int_0^(pi/2) sin^2(x) cos^2(x) dx` Evaluate the integral

Wednesday, October 12, 2016

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\sin}}^{{2}}{\left({x}\right)}{{\cos}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\sin}}^{{2}}{x}{{\cos}}^{{2}}{x}{\left.{d}{x}\right.}$

By squaring the formula for sine of double angle $\displaystyle{\sin{{2}}}\theta={2}{\sin{\theta}}{\cos{\theta}}$ we get $\displaystyle{{\sin}}^{{2}}{2}\theta={4}{{\sin}}^{{2}}\theta{{\cos}}^{{2}}\theta$

$\displaystyle\frac{{1}}{{4}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\sin}}^{{2}}{2}{x}{\left.{d}{x}\right.}$

Make substitution: $\displaystyle{t}={2}{x}\Rightarrow{\left.{d}{x}\right.}=\frac{{\left.{d}{t}}}{{2}}.$ New limits of integration are $t_1=2\cdot0=0$ and $\displaystyle{t}_{{2}}={2}\cdot\frac{\pi}{{2}}=\pi$

$\displaystyle\frac{{1}}{{8}}{\int_{{0}}^{\pi}}{{\sin}}^{{2}}{t}{\left.{d}{t}\right.}$

Use formula for cosine of double angle to obtain $\displaystyle{{\sin}}^{{2}}\theta=\frac{{{1}-{\cos{{2}}}\theta}}{{2}}$

$\displaystyle\frac{{1}}{{16}}{\int_{{0}}^{\pi}}{\left({1}-{\cos{{2}}}{t}\right)}{\left.{d}{t}\right.}$

Make substitution: $\displaystyle{u}={2}{t}\Rightarrow{\left.{d}{t}\right.}=\frac{{{d}{u}}}{{2}}.$ New limits of integration are $\displaystyle{u}_{{1}}={0}$ and $\displaystyle{u}_{{2}}={2}\pi.$

$\displaystyle\frac{{1}}{{32}}{\int_{{0}}^{{{2}\pi}}}{\left({1}-{\cos{{u}}}\right)}{d}{u}=\frac{{1}}{{32}}{\left({u}-{\sin{{u}}}\right)}{{\mid}_{{0}}^{{{2}\pi}}}=\frac{{1}}{{32}}{\left({2}\pi-{0}-{0}+{0}\right)}=\frac{\pi}{{16}}$

The integral is equal to $\displaystyle\frac{\pi}{{16}}.$

The Notes

Wednesday, October 12, 2016

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\sin}}^{{2}}{\left({x}\right)}{{\cos}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

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