Integration

This rule of integration 0af(x)dx=0af(ax)dx\displaystyle\int \limits^{a}_{0}f\left( x\right) dx =\displaystyle\int \limits^{a}_{0}f\left( a - x\right) dx I tried by using this rule to solve this integration but it wasn't solve.. the integration is 0πx2tan2(x)dx\displaystyle\int \limits^{\pi }_{0}\frac{x}{2 - \tan \nolimits^{2}\left( x\right) } dx ... I tried wolfram alpha also but it was useless ..... How we can solve this integration by using this rule? . Thanks

Note by David Danial
4 years ago

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you can apply the rule then add the new integral to the original one, and divide by 2. You will be left with the integral:π20π12tan2xdx\displaystyle \frac{\pi}{2} \int_0^{\pi} \frac{1}{2-\tan^2x} dx .

Did you try that??

Hasan Kassim - 4 years ago

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What will happen if we solve 2tan2x=02 - tan^{2} x = 0 x(0,π) x \in (0, \pi) ?.....

David Danial - 4 years ago

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yes there is discontinueties in the integral at :

arctan(2) \arctan (\sqrt{2} ) which is between π4 \frac{\pi}{4} and π2 \frac{\pi}{2} .

And πarctan(2) \pi - \arctan (\sqrt{2} ) .

Hasan Kassim - 4 years ago

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@Hasan Kassim So this integration is divergent.... could you please explain more this part of discontinuous.?

David Danial - 4 years ago

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@David Danial Yes the integral diverges: lets take take a part of it :

0arctan(2)dx2tan2x\displaystyle \int_0^{\arctan (\sqrt{2})} \frac{dx}{2-\tan^2x}

=limaarctan(2)0adx2tan2x\displaystyle = \lim_{a \to \arctan (\sqrt{2})} \int_0^{a} \frac{dx}{2-\tan^2x}

Use the substitution u=tanx u=\tan x :

=limaarctan(2)0tanadu(2u2)(1+u2)\displaystyle = \lim_{a \to \arctan (\sqrt{2})} \int_0^{\tan a } \frac{du}{(2-u^2)(1+u^2)}

Easy by applying partial fractions then integrating, then apply the limit and it will yield a value of infinty. So the whole Integral diverges.

Hasan Kassim - 4 years ago

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@Trevor B. You might like this!

Calvin Lin Staff - 4 years ago

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Yes, the Bounds Trick works nicely here. I know how to get to the indefinite, I just need to make the time to write it down and check the bounds. Partial fracs with what Hasan said originally should make it work.

Trevor B. - 4 years ago

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you could try with partial fractions too

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