Deragz

2022-01-01

I need help calculating two integrals
1) ${\int }_{1}^{2}\sqrt{4+\frac{1}{x}}dx$
2) ${\int }_{0}^{\frac{2}{\pi }}{x}^{n}\mathrm{sin}\left(x\right)dx$

lovagwb

For the first problem:
$\int \sqrt{4+\frac{1}{x}}dx=\int \frac{\sqrt{4x+1}}{\sqrt{x}}dx$
Let
$\int \frac{\sqrt{4{u}^{2}+1}}{u}2udu$
Try some trigonometric substitution.

Lindsey Gamble

For the second problem:
Set
$I\left(n\right)=\int {x}^{n}\mathrm{sin}\left(x\right)dx$
Let by integration be parts.
$I\left(n\right)={x}^{n}×\mathrm{cos}x+n\int {x}^{n-1}\mathrm{cos}xdx$
Let by integration be parts.
$I\left(n\right)={x}^{n}×\mathrm{cos}x+n\left({x}^{n-1}\mathrm{sin}x-\left(n-1\right)\int {x}^{n-2}\mathrm{sin}xdx\right)={x}^{n}×\mathrm{cos}x+n{x}^{n-1}\mathrm{sin}x-n\left(n-1\right)I\left(n-2\right)$
Then use mathematic induction you will get the formula.

karton

You can indeed work out the second integral by parts.
${I}_{n}:={\int }_{0}^{\pi /2}{x}^{n}\mathrm{sin}xdx=-{x}^{n}\mathrm{cos}x{|}_{0}^{\pi /2}+n{\int }_{0}^{\pi /2}{x}^{n-1}\mathrm{cos}xdx$
Repeat with the cosine integral,
${J}_{n}:={\int }_{0}^{\pi /2}{x}^{n}\mathrm{cos}xdx={x}^{n}\mathrm{sin}x{|}_{0}^{\pi /2}-n{\int }_{0}^{\pi /2}{x}^{n-1}\mathrm{sin}xdx$
This gives you the recurrence relations
${I}_{n}=n{J}_{n-1}$
${J}_{n}=\left(\frac{\pi }{2}{\right)}^{n}-n{I}_{n-1}$
So that
${I}_{n}=n\left(\left(\frac{\pi }{2}{\right)}^{n-1}-\left(n-1\right){I}_{n-2}\right)$
When you decrease n, you will eventually reach n=1 or n=0, you need to explictly compute ${I}_{1}$ and ${I}_{0}$. It is an easy matter to establish
${I}_{0}={J}_{0}=1$
then
${I}_{1}=1$

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