Trigonometric Definite Integral (1 Viewer)

juantheron · Jun 26, 2014

$Evaluation of $\displaystyle \int_{0}^{\pi}x\cdot \cot x dx$

$I have Tried like this way.$

$Here $\displaystyle \cot x$ is $0$at $\displaystyle x = \frac{\pi}{2}$ .

$So we can break integral$

$as $\displaystyle \int_{0}^{\frac{\pi}{2}}x\cdot \cot xdx +\int_{\frac{\pi}{2}}^{\pi}x\cdot \cot x dx$

$I have calculated first part and getting $\displaystyle \frac{\pi}{2}\ln(2)$ but did not understand$

$How can I calculate second Integral$

Help me

Thanks

Thanks carrotsticks I have edited it.

Carrotsticks · Jun 26, 2014

cot(x) is not undefined at x=pi/2, that's tan(x).

Squar3root · Jun 26, 2014

Use parts

Carrotsticks · Jun 26, 2014

Squar3root said:
Use parts

The integral diverges, by parts would be a lot more difficult.

$\\ \int_{\frac{\pi}{2}}^{\pi} x \cot x \ dx < \int_{\frac{\pi}{2}}^{\pi} \cot x \ dx $ since $ x \cot x \leq 0 $ for all $ x \in \left [ \frac{\pi}{2},\pi \right ]$

So by the comparison test, the left integral diverges to negative infinity since the right integral diverges to negative infinity.

juantheron · Jun 26, 2014

Thanks Square3root Using Yours Hint.

$Using Integration by parts for second$

$Let $\displaystyle I = \int_{\frac{\pi}{2}}^{\pi}x\cdot \cot xdx = \left[x\cdot \ln(\sin x)\right]_{\frac{\pi}{2}}^{\pi}-\int_{\frac{\pi}{2}}^{\pi}\ln (\sin x)dx$

Now How can I solve after that

seanieg89 · Jun 26, 2014

juantheron said:
Thanks Square3root Using Yours Hint.

$Using Integration by parts for second$

$Let $\displaystyle I = \int_{\frac{\pi}{2}}^{\pi}x\cdot \cot xdx = \left[x\cdot \ln(\sin x)\right]_{\frac{\pi}{2}}^{\pi}-\int_{\frac{\pi}{2}}^{\pi}\ln (\sin x)dx$

Now How can I solve after that

Why are you still trying to solve this? Carrot showed that the integral does not converge.

Trigonometric Definite Integral (1 Viewer)

juantheron

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Carrotsticks

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Squar3root

realest nigga

Carrotsticks

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juantheron

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seanieg89

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