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I have no idea how to go about solving this integral: $$ \int_0^{\frac{\pi}{2}-}\frac{dx}{1+\tan^{\sqrt{2}}\left(x\right)} .$$ The most I've come up with is rewriting it as a series to obtain $$ \int_0^{\frac{\pi}{2}-}\sum\limits_{n=0}^\infty \left\{-\tan^{\sqrt{2}}\left(x\right)\right\}^n{dx}, $$ but then I realized that this resulted in a hideous set of integrations if I were to try integrating term-by-term and then turning the answer back: $$ \int_0^{\frac{\pi}{2}-}\left[1-\tan^{\sqrt{2}}\left(x\right)+\tan^2 (x)-\tan^{2\cdot\sqrt{2}}(x)+\cdots\right]\,dx, $$ because then I couldn't solve $$ \int \tan^\sqrt{2} (x) \, dx. $$ So to cope with the above I thought I might expand even further, but then I realized I must be waaayyyyy off track from where I should be.

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  • $\begingroup$ $$\frac1{1+ \tan^{\sqrt2}x}=\frac{\cos^{\sqrt2}x}{\sin^{\sqrt2}x+\cos^{\sqrt2}x}$$ Then math.stackexchange.com/questions/439851/… $\endgroup$ Sep 28 '14 at 5:18
  • $\begingroup$ Thanks for this, I've spotted that relation already but I hadn't thought of what they did in the link you've provided, which is really interesting. I'll give it a shot and see what happens. $\endgroup$
    – bjd2385
    Sep 28 '14 at 6:06
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I suggest you use the substitution $x=\frac{\pi}{2}-u$. Then use the identity $\tan(\frac{\pi}{2}-u)=\cot(u)$. Now look at the two different representations of the integral you have and add them.

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$$\int_{0}^{\frac{\pi}{2}-}\frac{dx}{1+{tan}^{\sqrt{2}}\left(x\right)} = \int_{0}^{\frac{\pi}{2}}\frac{{cos}^{\sqrt{2}}\left(x\right)\:dx}{{cos}^{\sqrt{2}}\left(x\right)+ {sin}^{\sqrt{2}}\left(x\right)},$$ $$ \int_{0}^{\frac{\pi}{2}-}\frac{dx}{1+{tan}^{\sqrt{2}}\left(x\right)}= \int_{0}^{\frac{\pi}{2}}\frac{{sin}^{\sqrt{2}}\left(x\right)\:dx}{{sin}^{\sqrt{2}}\left(x\right)+ {cos}^{\sqrt{2}}\left(x\right)},$$ $$\therefore \int_{0}^{\frac{\pi}{2}}\left\{\frac{{cos}^{\sqrt{2}}\left(x\right)}{{cos}^{\sqrt{2}}\left(x\right)+{sin}^{\sqrt{2}}\left(x\right)}+\frac{{sin}^{\sqrt{2}}\left(x\right)}{{sin}^{\sqrt{2}}\left(x\right)+ {cos}^{\sqrt{2}}\left(x\right)}\right\}dx = \int_{0}^{\frac{\pi}{2}}dx=\frac{\pi}{2},$$

which is indeed twice the expected result because we added the two integrals together, so this must be divided by two to get $$ \int_0^{\frac{\pi}{2}-}\frac{dx}{1+{tan}^{\sqrt{2}}\left(x\right)} = \boxed{\frac{\pi}{4}.} $$

Can be expanded to solve integrals of the form

$$ \int_{0}^{\frac{\pi}{2}}\frac{{cos}^{n}\left(x\right)\:dx}{{cos}^{n}\left(x\right)+ {sin}^{n}\left(x\right)} .$$

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