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# 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
1 year, 7 months ago

## Comments

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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?? · 1 year, 7 months ago

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What will happen if we solve $$2 - tan^{2} x = 0$$ $$x \in (0, \pi)$$?..... · 1 year, 7 months ago

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yes there is discontinueties in the integral at :

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

And $$\pi - \arctan (\sqrt{2} )$$. · 1 year, 7 months ago

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So this integration is divergent.... could you please explain more this part of discontinuous.? · 1 year, 7 months ago

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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. · 1 year, 7 months ago

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you could try with partial fractions too · 1 year, 7 months ago

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@Trevor B. You might like this! Staff · 1 year, 7 months 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. · 1 year, 7 months ago

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