The Notes: `int_0^(pi/2) (2 - sin(theta))^2 d theta` Evaluate the integral

Wednesday, June 3, 2015

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

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\left({2}-{\sin{{\left(\theta\right)}}}\right)}}^{{2}}{d}\theta$

First, expand the integrand.

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left({2}-{\sin{{\left(\theta\right)}}}\right)}{\left({2}-{\sin{{\left(\theta\right)}}}\right)}{d}\theta$

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left({4}-{4}{\sin{{\left(\theta\right)}}}+{{\sin}}^{{2}}{\left(\theta\right)}\right)}{d}\theta$

For the last term of the integrand, apply the trigonometric identity $\displaystyle{\cos{{\left({2}\theta\right)}}}={1}-{2}{{\sin}}^{{2}}{\left(\theta\right)}$ .

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left({4}-{4}{\sin{{\left(\theta\right)}}}+\frac{{1}}{{2}}-\frac{{\cos{{\left({2}\theta\right)}}}}{{2}}\right)}{d}\theta$

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left(\frac{{9}}{{2}}-{4}{\sin{\theta}}-\frac{{{\cos{{\left({2}\theta\right)}}}}}{{2}}\right)}{d}\theta$

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{9}}{{2}}{d}\theta-{\int_{{0}}^{{\frac{\pi}{{2}}}}}{4}{\sin{\theta}}{d}\theta-{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{{\cos{{\left({2}\theta\right)}}}}}{{2}}{d}\theta$

Apply the integral formulas $\displaystyle\int{a}{\left.{d}{x}\right.}={a}{x}$ , $\displaystyle\int{\sin{{x}}}{\left.{d}{x}\right.}=-{\cos{{x}}}$ and $\displaystyle\int{\cos{{x}}}{\left.{d}{x}\right.}={\sin{{x}}}$ .

$\displaystyle={\left(\frac{{9}}{{2}}\theta+{4}{\cos{{\left(\theta\right)}}}-\frac{{{\sin{{\left({2}\theta\right)}}}}}{{4}}\right)}{{\mid}_{{0}}^{{\frac{\pi}{{2}}}}}$

$\displaystyle={\left(\frac{{9}}{{2}}\cdot\frac{\pi}{{2}}+{4}{\cos{{\left(\frac{\pi}{{2}}\right)}}}-\frac{{\sin{{\left({2}\cdot\frac{\pi}{{2}}\right)}}}}{{4}}\right)}-{\left(\frac{{9}}{{2}}\cdot{0}+{4}{\cos{{\left({0}\right)}}}-\frac{{{\sin{{\left({2}\cdot{0}\right)}}}}}{{4}}\right)}$

$\displaystyle={\left(\frac{{{9}\pi}}{{4}}+{4}\cdot{0}+\frac{{0}}{{4}}\right)}-{\left({0}+{4}\cdot{1}-\frac{{0}}{{4}}\right)}$

$\displaystyle=\frac{{{9}\pi}}{{4}}-{4}$

Therefore, $\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\left({2}-{\sin{{\left(\theta\right)}}}\right)}}^{{2}}{d}\theta=\frac{{{9}\pi}}{{4}}-{4}$ .

The Notes

Wednesday, June 3, 2015

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

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