# Evaluate the following definite integral. \int_{0}^{\frac{\pi}{2}}x \cos 2x dx

Question
Applications of integrals
Evaluate the following definite integral.
$$\displaystyle{\int_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}{x}{\cos{{2}}}{x}{\left.{d}{x}\right.}$$

2021-03-06
$$\displaystyle{\int_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}{x}{\cos{{2}}}{x}{\left.{d}{x}\right.}$$
Integrate by parts:
$$\displaystyle\int{f}{g}'={f}{g}-\int{f}'{g}$$
f=x, f'=1
$$\displaystyle{g}'={\cos{{2}}}{x},{g}={\frac{{{\sin{{2}}}{x}}}{{{2}}}}$$
$$\displaystyle\Rightarrow{{\left[{\frac{{{x}{\sin{{\left({2}{x}\right)}}}}}{{{2}}}}\right]}_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}-{\int_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}{\frac{{{\sin{{\left({2}{x}\right)}}}}}{{{2}}}}{\left.{d}{x}\right.}$$
Solve
$$\displaystyle{\int_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}{\frac{{{\sin{{2}}}{x}}}{{{2}}}}{\left.{d}{x}\right.}$$
Sunstitute u =2x
$$\displaystyle{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}={2}$$
$$\displaystyle{\left.{d}{x}\right.}={\frac{{{1}}}{{{2}}}}{d}{u}$$
$$\displaystyle={\frac{{{1}}}{{{4}}}}{\int_{{{0}}}^{{\pi}}}{\sin{{\left({u}\right)}}}{d}{u}$$
Now solve $$\displaystyle{\int_{{{0}}}^{{\pi}}}{\sin{{u}}}{d}{u}$$
This is a standard integral: $$\displaystyle={{\left[-{\cos{{u}}}\right]}_{{{0}}}^{{\pi}}}$$
Plug in the solved integrals:
$$\displaystyle{\frac{{{1}}}{{{2}}}}{\int_{{{0}}}^{{\pi}}}{\sin{{u}}}{d}{u}={{\left[{\frac{{-{\cos{{\left({u}\right)}}}}}{{{4}}}}\right]}_{{{0}}}^{{\pi}}}$$
Undo substitution u=2x
$$\displaystyle={{\left[{\frac{{-{\cos{{\left({2}{x}\right)}}}}}{{{4}}}}\right]}_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}$$
Plug in the solved integrals:
$$\displaystyle{{\left[{\frac{{{x}{\sin{{\left({2}{x}\right)}}}}}{{{2}}}}\right]}_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}-{\int_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}{\frac{{{\sin{{\left({2}{x}\right)}}}}}{{{2}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={{\left[{\frac{{{x}{\sin{{\left({2}{x}\right)}}}}}{{{2}}}}+{\frac{{{\cos{{\left({2}{x}\right)}}}}}{{{4}}}}\right]}_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}$$
$$\displaystyle={\left[{\frac{{{\frac{{\pi}}{{{2}}}}{\sin{{\left({2}.{\frac{{\pi}}{{{2}}}}\right)}}}}}{{{2}}}}+{\frac{{{\cos{{\left({2}.{\frac{{\pi}}{{{2}}}}\right)}}}}}{{{4}}}}\right]}-{\left[{\frac{{{0}{\sin{{\left({2}{x}{0}\right)}}}}}{{{2}}}}+{\frac{{{\cos{{\left({2.0}\right)}}}}}{{{4}}}}\right]}$$
$$\displaystyle={\left[{\frac{{{\frac{{\pi}}{{{2}}}}{\sin{{\left(\pi\right)}}}}}{{{2}}}}+{\frac{{{\cos{{\left(\pi\right)}}}}}{{{4}}}}\right]}-{\left[{0}+{\frac{{{\cos{{\left({0}\right)}}}}}{{{4}}}}\right]}$$
$$\displaystyle={\left[{0}-{\frac{{{1}}}{{{4}}}}\right]}-{\left[{0}+{\frac{{{1}}}{{{4}}}}\right]}$$
$$\displaystyle=-{\frac{{{1}}}{{{4}}}}-{\frac{{{1}}}{{{4}}}}$$
$$\displaystyle={\frac{{-{1}-{1}}}{{{4}}}}$$
$$\displaystyle={\frac{{-{2}}}{{{4}}}}$$
$$\displaystyle={\frac{{-{1}}}{{{2}}}}$$

### Relevant Questions

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