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In this question sos440 has mentioned about an integral that he computed:

$$\int_0^{\pi/2}\arctan(1-\sin^2 (x) \cos^2 (x))dx = \pi \left( \frac{\pi}{4}- \arctan \sqrt{\frac{\sqrt{2}-1}{2}}\right)$$

I would really like to know how this can be proved. I tried using differentiation under the integral sign on parmeter $a$:

$$\int_0^{\pi/2}\arctan(1-a\sin^2 (x) \cos^2 (x))dx $$

but it didn't really work. I would be really grateful if sos440 or somebody else can tell me the secret behind discovering this formula.

Thanks!

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  • $\begingroup$ @sos440: Thank You so much! I like your blog. $\endgroup$ – Anthony Mar 28 '13 at 8:18
  • $\begingroup$ @sos440 why don't you post this comment as answer? $\endgroup$ – Norbert Mar 28 '13 at 8:57
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Following Norbert's advice, I deleted my comment and moved it to an answer:

Complex analysis may help you: see my blog posting.

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  • $\begingroup$ Sir, my chrome browser can't open this page. $\endgroup$ – r9m Jun 28 '17 at 1:37
  • $\begingroup$ @r9m It seems that the hosting company somehow lost all the images and unfortunately I forgot how to solve this. Thankfully the user Pranav Arora has a nice solution. $\endgroup$ – Sangchul Lee Jun 28 '17 at 13:17
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$$ \begin{aligned} \int_0^{\pi/2} \arctan\left(1-\cos^2x\sin^2x\right)\,dx & =\int_0^{\pi/2} \left(\frac{\pi}{2}-\arctan\left(\frac{1}{1-\sin^2x\cos^2x}\right)\right)\,dx \\ &=\frac{\pi^2}{4}-\int_0^{\pi/2} \arctan(\sin^2x)\,dx-\int_0^{\pi/2} \arctan(\cos^2x)\,dx \\ &=\frac{\pi^2}{4}-2\int_0^{\pi/2}\arctan(\cos^2x)\,dx \\ \end{aligned} $$

Consider

$$I(a)=\int_0^{\pi/2} \arctan(a\,\cos^2x)\,dx $$

$$ \begin{aligned} \Rightarrow I'(a)&=\int_0^{\pi/2} \frac{\cos^2x}{1+a^2\cos^4x}\,dx \\ &= \int_0^{\pi/2} \frac{\sec^2x}{\sec^4x+a^2}\,dx\\ &= \int_0^{\pi/2} \frac{\sec^2x}{(1+\tan^2x)^2+a^2}\,dx\\ &=\int_0^{\infty} \frac{dt}{t^4+2t^2+a^2+1}\,\,\,\,\,\,\,(\tan x=t) \\ \end{aligned} $$

With the substitution $t\mapsto \dfrac{\sqrt{a^2+1}}{t} $,

$$I'(a)=\frac{1}{\sqrt{a^2+1}}\int_0^{\infty} \frac{t^2}{t^4+2t^2+a^2+1}\,dt $$

$$ \begin{aligned} \Rightarrow I'(a) &=\frac{1}{2\sqrt{a^2+1}}\int_0^{\infty} \frac{\sqrt{a^2+1}+t^2}{t^4+2t^2+a^2+1}\,dt \\ &=\frac{1}{2\sqrt{a^2+1}}\int_0^{\infty} \frac{1+\frac{\sqrt{1+a^2}}{t^2}}{\left(t-\frac{\sqrt{a^2+1}}{t}\right)^2+2(1+\sqrt{a^2+1})}\,dt\\ &=\frac{1}{2\sqrt{a^2+1}}\int_{-\infty}^{\infty} \frac{dy}{y^2+2(1+\sqrt{a^2+1})}\,\,\,\,\,\,\,\left(t-\frac{\sqrt{a^2+1}}{t}=y\right) \\ &=\frac{\pi}{2\sqrt{2}} \frac{1}{\sqrt{1+a^2}\sqrt{1+\sqrt{a^2+1}}} \\ \end{aligned} $$

Integrating back,

$$ \begin{aligned} I(1)-I(0)=I(1) &=\frac{\pi}{2\sqrt{2}}\int_0^1 \frac{da}{\sqrt{1+a^2}\sqrt{1+\sqrt{a^2+1}}} \\ &= \frac{\pi}{2\sqrt{2}}\int_1^{\sqrt{2}} \frac{dt}{\sqrt{t-1}(t+1)}\,\,\,\,\,\,\,(\sqrt{a^2+1}=t) \\ &=\frac{\pi}{\sqrt{2}}\int_0^{\sqrt{\sqrt{2}-1}} \frac{du}{u^2+2} \,\,\,\,\,\,\,(t-1=u^2) \\ &=\frac{\pi}{2} \left(\arctan\frac{u}{\sqrt{2}}\right|_0^{\sqrt{\sqrt{2}-1}} \\ &=\frac{\pi}{2}\arctan\left(\sqrt{\frac{\sqrt{2}-1}{2}}\right) \\ \end{aligned}$$

Hence,

$$\boxed{\displaystyle \int_0^{\pi/2} \arctan\left(1-\cos^2x\sin^2x\right)\,dx=\dfrac{\pi^2}{4}-\pi\arctan\left(\sqrt{\dfrac{\sqrt{2}-1}{2}}\right)}$$

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  • $\begingroup$ Very clever use of the sum identity for arctan. +1 $\endgroup$ – Random Variable Dec 13 '14 at 3:30
  • $\begingroup$ @RandomVariable; Thank you! :) $\endgroup$ – Pranav Arora Dec 13 '14 at 5:08
  • $\begingroup$ This is well done and great! $\endgroup$ – Madona Syombua Jun 22 '15 at 6:32
  • $\begingroup$ @Madona: Thanks! :) $\endgroup$ – Pranav Arora Jun 22 '15 at 6:51
  • $\begingroup$ @PranavArora How did you get the $t\mapsto \dfrac{\sqrt{a^2+1}}{t}$ substitution? $\endgroup$ – Kugelblitz Jun 17 '17 at 5:16

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