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

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

Did you try that?? – Hasan Kassim · 1 year, 11 months ago

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– David Danial · 1 year, 11 months ago

What will happen if we solve \(2 - tan^{2} x = 0 \) \( x \in (0, \pi) \)?.....Log in to reply

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

And \( \pi - \arctan (\sqrt{2} ) \). – Hasan Kassim · 1 year, 11 months ago

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– David Danial · 1 year, 11 months ago

So this integration is divergent.... could you please explain more this part of discontinuous.?Log in to reply

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

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

Use the substitution \( u=\tan x \):

\[\displaystyle = \lim_{a \to \arctan (\sqrt{2})} \int_0^{\tan a } \frac{du}{(2-u^2)(1+u^2)} \]

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 · 1 year, 11 months ago

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you could try with partial fractions too – Hjalmar Orellana Soto · 1 year, 11 months ago

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@Trevor B. You might like this! – Calvin Lin Staff · 1 year, 11 months ago

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– Trevor B. · 1 year, 11 months ago

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.Log in to reply