# Integration

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 Note by David Danial
5 years, 2 months 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:$\displaystyle \frac{\pi}{2} \int_0^{\pi} \frac{1}{2-\tan^2x} dx$.

Did you try that??

- 5 years, 2 months ago

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

- 5 years, 2 months ago

yes there is discontinueties in the integral at :

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

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

- 5 years, 2 months ago

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

- 5 years, 2 months ago

Yes the integral diverges: lets take take a part of it :

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

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

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.

- 5 years, 2 months ago

@Trevor B. You might like this!

Staff - 5 years, 2 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.

- 5 years, 2 months ago

you could try with partial fractions too

- 5 years, 2 months ago