Stefan Hendricks

2021-12-29

Use a table of integrals to evaluate the definite integral.
${\int }_{0}^{\frac{\pi }{2}}x\mathrm{sin}2xdx$

Philip Williams

Step 1
Given: $I={\int }_{0}^{\frac{\pi }{2}}x\mathrm{sin}\left(2x\right)dx$
for evaluating given integral, we use integral by parts theorem
according to this theorem
$\int \left({f}^{\prime }\left(x\right)\right)g\left(x\right)dx=g\left(x\right)\int {f}^{\prime }\left(x\right)dx-\int \left[\left({g}^{\prime }\left(x\right)\right)\int {f}^{\prime }\left(x\right)dx\right]dx+c$
Step 2
so,
${\int }_{0}^{\frac{\pi }{2}}x\mathrm{sin}\left(2x\right)={\left[x\int \mathrm{sin}2xdx-\int \left[1\int \mathrm{sin}2xdx\right]dx\right]}_{0}^{\frac{\pi }{2}}$
$\left(\because \int \mathrm{sin}kxdx=-\frac{\mathrm{cos}kx}{k}+c\right)$
$={\left[x\left(-\frac{\mathrm{cos}2x}{2}\right)-\int \left(-\frac{\mathrm{cos}2x}{2}\right)dx\right]}_{0}^{\frac{\pi }{2}}$
$\left(\because \int \mathrm{cos}kxdx=\frac{\mathrm{sin}kx}{k}+c\right)$
$={\left[-\frac{x\mathrm{cos}2x}{2}+\frac{1}{2}\frac{\mathrm{sin}2x}{2}\right]}_{0}^{\frac{\pi }{2}}$
$={\left[\frac{-2x\mathrm{cos}2x+\mathrm{sin}2x}{4}\right]}_{0}^{\frac{\pi }{2}}$
$=\frac{1}{4}\left[\left(-2\left(\frac{\pi }{2}\right)\mathrm{cos}\left(\pi \right)+\mathrm{sin}\left(\pi \right)\right)-\left(-2\left(0\right)\mathrm{cos}\left(0\right)+\mathrm{sin}\left(0\right)\right)\right]$
$=\frac{1}{4}\left[-\pi \left(-1\right)+0+0-0\right]$
$=\frac{1}{4}\left(\pi \right)$
$=\frac{\pi }{4}$
hence, given integral is equal to

veiga34

$\int x\mathrm{sin}\left(2x\right)dx$
$=-\frac{x\mathrm{cos}\left(2x\right)}{2}-\int -\frac{\mathrm{cos}\left(2x\right)}{2}dx$
$\int -\frac{\mathrm{cos}\left(2x\right)}{2}dx$
$=-\frac{1}{4}\int \mathrm{cos}\left(u\right)du$
$\int \mathrm{cos}\left(u\right)du$
$=\mathrm{sin}\left(u\right)$
$-\frac{1}{4}\int \mathrm{cos}\left(u\right)du$
$=-\frac{\mathrm{sin}\left(u\right)}{4}$
$=-\frac{\mathrm{sin}\left(2x\right)}{4}$
$-\frac{x\mathrm{cos}\left(2x\right)}{2}-\int -\frac{\mathrm{cos}\left(2x\right)}{2}dx$
$=\frac{\mathrm{sin}\left(2x\right)}{4}-\frac{x\mathrm{cos}\left(2x\right)}{2}$
$\int x\mathrm{sin}\left(2x\right)dx$
$=\frac{\mathrm{sin}\left(2x\right)}{4}-\frac{x\mathrm{cos}\left(2x\right)}{2}+C$
$=\frac{\mathrm{sin}\left(2x\right)-2x\mathrm{cos}\left(2x\right)}{4}+C$

karton

${\int }_{0}^{\frac{\pi }{2}}x×\mathrm{sin}\left(2x\right)dx$
Evaluate the indefinite integral
$\begin{array}{}x×\mathrm{sin}\left(2x\right)dx\\ \text{Use the partial integration}\\ \\ x×\left(-\frac{\mathrm{cos}\left(2x\right)}{2}\right)-\int -\frac{\mathrm{cos}\left(2x\right)}{2}dx\\ x×\left(-\frac{\mathrm{cos}\left(2x\right)}{2}\right)-1×\left(-\frac{1}{2}\right)×\int \mathrm{cos}\left(2x\right)dx\\ x×\left(-\frac{\mathrm{cos}\left(2x\right)}{2}\right)+\frac{1}{2}×\int \frac{\mathrm{cos}\left(t\right)}{2}dt\\ x×\left(-\frac{\mathrm{cos}\left(2x\right)}{2}\right)+\frac{1}{2}×\frac{1}{2}×\int \mathrm{cos}\left(t\right)dt\\ x×\left(-\frac{\mathrm{cos}\left(2x\right)}{2}\right)+\frac{1}{4}×\mathrm{sin}\left(t\right)\\ x×\left(-\frac{\mathrm{cos}\left(2x\right)}{2}\right)+\frac{1}{4}×\mathrm{sin}\left(2x\right)\\ -\frac{x×\mathrm{cos}\left(2x\right)}{2}+\frac{\mathrm{sin}\left(2x\right)}{4}\\ \left(-\frac{x×\mathrm{cos}\left(2x\right)}{2}+\frac{\mathrm{sin}\left(2x\right)}{4}\right){|}_{0}^{\frac{\pi }{2}}\\ -\frac{\frac{\pi }{2}×\mathrm{cos}\left(2×\frac{\pi }{2}\right)}{2}+\frac{\mathrm{sin}\left(2×\frac{\pi }{2}\right)}{4}-\left(-\frac{0\mathrm{cos}\left(2×0\right)}{2}+\frac{\mathrm{sin}\left(2×0\right)}{4}\right)\end{array}$
Simplify
Solution
$\frac{\pi }{4}$

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