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

Question
Applications of integrals
asked 2021-03-05
Evaluate the following definite integral.
\(\displaystyle{\int_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}{x}{\cos{{2}}}{x}{\left.{d}{x}\right.}\)

Answers (1)

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:
\(\displaystyle{{\left[{\frac{{{x}{\sin{{\left({2}{x}\right)}}}}}{{{2}}}}\right]}_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}-{\int_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}{\frac{{{\sin{{\left({2}{x}\right)}}}}}{{{2}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={{\left[{\frac{{{x}{\sin{{\left({2}{x}\right)}}}}}{{{2}}}}+{\frac{{{\cos{{\left({2}{x}\right)}}}}}{{{4}}}}\right]}_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}\)
\(\displaystyle={\left[{\frac{{{\frac{{\pi}}{{{2}}}}{\sin{{\left({2}.{\frac{{\pi}}{{{2}}}}\right)}}}}}{{{2}}}}+{\frac{{{\cos{{\left({2}.{\frac{{\pi}}{{{2}}}}\right)}}}}}{{{4}}}}\right]}-{\left[{\frac{{{0}{\sin{{\left({2}{x}{0}\right)}}}}}{{{2}}}}+{\frac{{{\cos{{\left({2.0}\right)}}}}}{{{4}}}}\right]}\)
\(\displaystyle={\left[{\frac{{{\frac{{\pi}}{{{2}}}}{\sin{{\left(\pi\right)}}}}}{{{2}}}}+{\frac{{{\cos{{\left(\pi\right)}}}}}{{{4}}}}\right]}-{\left[{0}+{\frac{{{\cos{{\left({0}\right)}}}}}{{{4}}}}\right]}\)
\(\displaystyle={\left[{0}-{\frac{{{1}}}{{{4}}}}\right]}-{\left[{0}+{\frac{{{1}}}{{{4}}}}\right]}\)
\(\displaystyle=-{\frac{{{1}}}{{{4}}}}-{\frac{{{1}}}{{{4}}}}\)
\(\displaystyle={\frac{{-{1}-{1}}}{{{4}}}}\)
\(\displaystyle={\frac{{-{2}}}{{{4}}}}\)
\(\displaystyle={\frac{{-{1}}}{{{2}}}}\)
0

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