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Find $\displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{1}{1+\tan(x)^{\sqrt{2}}}dx$.

I am told I should solve this by switching the limits, but I am unsure of the substitution I should use.

Olivier Oloa
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Puzzled417
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2 Answers2

1

You may observe that, by the change of variable $x \to \pi/2-x$, you get $$I=\int_{0}^{\dfrac{\pi}{2}} \dfrac{1}{1+\tan(x)^{\sqrt{2}}}dx=\int_{0}^{\dfrac{\pi}{2}} \dfrac{1}{1+\frac1{\tan(x)^{\sqrt{2}}}}dx=\int_{0}^{\dfrac{\pi}{2}} \dfrac{\tan(x)^{\sqrt{2}}}{1+\tan(x)^{\sqrt{2}}}dx $$ giving $$ 2I=I+I=\int_{0}^{\dfrac{\pi}{2}} \dfrac{1}{1+\tan(x)^{\sqrt{2}}}dx+\int_{0}^{\dfrac{\pi}{2}} \dfrac{\tan(x)^{\sqrt{2}}}{1+\tan(x)^{\sqrt{2}}}dx=\int_{0}^{\dfrac{\pi}{2}} dx=\frac{\pi}{2} $$ or $$ I=\color{blue}{\frac{\pi}4}. $$

Olivier Oloa
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  • Substitutions are typically taught using variable other than that being integrated with respect to. Thus this answer would be easier to understand if it said let $u = \frac{\pi}{2} - x$ – Tanner Carawan Dec 19 '15 at 23:15
  • @TannerCarawan Then $u$ is a new "dummy" variable, which we can change back to $x$. So, we save ourselves a step by enforcing the substitution $x\mapsto \pi/2 - x$. – Mark Viola Dec 20 '15 at 00:32
  • Olivier. First, Happy New Year. Now, I am just curious. I posted my answer to this question nearly one month ago. And today, my answer received two down votes? Did you experience something similar? - Mark – Mark Viola Jan 17 '16 at 18:18
  • @Dr.MV Thanks. Happy New Year to you. "Did you experience something similar?": yes. Please have a look at my reputation points. – Olivier Oloa Jan 17 '16 at 19:08
  • If both you and I received multiple down votes today, then it makes me wonder if there is a system error. – Mark Viola Jan 17 '16 at 19:20
  • It's not only us. I thought it was someone with some malicious intent... – Olivier Oloa Jan 17 '16 at 19:30
1

Let $I(\alpha)$ be given by

$$I(\alpha)=\int_0^{\pi/2}\frac{1}{1+\tan^{\alpha}x}\,dx \tag 1$$

Then, let $x \mapsto \pi/2-x$ so that $\tan (\pi/2 -x)\mapsto \cot x$, $dx \mapsto -dx$, and the limits of integration go from $\pi/2$ to $0$. Then we have

$$I(\alpha) = \int_0^{\pi/2}\frac{\tan^{\alpha }x}{1+\tan^{\alpha }x}\,dx \tag 2$$

Adding $(1)$ and $(2)$ and dividing by two reveals that

$$I(\alpha)=\frac12\int_0^{\pi/2} \frac{1+\tan^{\alpha}x}{1+\tan^{\alpha}x}\,dx=\frac{\pi}{4}$$

which is independent of $\alpha$!

Mark Viola
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