Who can handle this integral ?

0π2arctan(1cos2xsin2x) dx=π24πarctan212. \int\limits_0^{\frac{\pi}{2}}\arctan\left(1-\cos^2 x\sin^2 x\right)\ \mathrm{d}x=\frac{\pi^2}{4}-\pi\arctan\sqrt{\frac{\sqrt{2}-1}{2}}. Can someone prove this ?

Note by Haroun Meghaichi
4 years, 11 months ago

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0π/2arctan(1cos2xsin2x)dx=0π/2(π2arctan(11sin2xcos2x))dx=π240π/2arctan(sin2x)dx0π/2arctan(cos2x)dx=π2420π/2arctan(cos2x)dx\displaystyle \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)=0π/2arctan(acos2x)dx\displaystyle I(a)=\int_0^{\pi/2} \arctan(a\,\cos^2x)\,dx

I(a)=0π/2cos2x1+a2cos4xdx=0π/2sec2xsec4x+a2dx=0π/2sec2x(1+tan2x)2+adx=0dtt4+2t2+a2+1(tanx=t)\displaystyle \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}\,dx\\ &=\int_0^{\infty} \frac{dt}{t^4+2t^2+a^2+1}\,\,\,\,\,\,\,(\tan x=t) \\ \end{aligned}

With the substitution ta2+1tt\mapsto \dfrac{\sqrt{a^2+1}}{t} ,

I(a)=1a2+10t2t4+2t2+a2+1dt\displaystyle I'(a)=\frac{1}{\sqrt{a^2+1}}\int_0^{\infty} \frac{t^2}{t^4+2t^2+a^2+1}\,dt

I(a)=12a2+10a2+1+t2t4+2t2+a2+1dt=12a2+101+1+a2t2(ta2+1t)2+2(1+a2+1)dt=12a2+1dyy2+2(1+a2+1)(ta2+1t=y)=π2211+a21+a2+1\displaystyle \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,

I(1)I(0)=I(1)=π2201da1+a21+a2+1=π2212dtt1(t+1)(a2+1=t)=π2021duu2+2(t1=u2)=π2(arctanu2021=π2arctan(212)\displaystyle \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,

0π/2arctan(1cos2xsin2x)dx=π24πarctan(212)\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)}

Pranav Arora - 4 years, 11 months ago

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What a great job! Hats off to you @Pranav Arora .

Haroun Meghaichi - 4 years, 11 months ago

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Wonderful solution Pranav!!! How do you think of such things?!¨\ddot\smile

Karthik Kannan - 4 years, 11 months ago

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Thank you Karthik! :)

I don't think I did anything special in this one, I used only the standard techniques as you can see. I am sure you too would have solved it if you jumped on this one before me. ;)

Pranav Arora - 4 years, 11 months ago

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@Pranav Arora Oh, no...I tried a lot but did not succeed. My approach was also 'differentiating under the integral' but I was nowhere close to the solution. Great solution once again!¨\ddot\smile

Karthik Kannan - 4 years, 11 months ago

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Excellent!!!

Parth Lohomi - 4 years, 10 months ago

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What level calculus is this..?I am in 12th grade am I supposed to be able to solve this...??

Rajsuryan Singh - 4 years, 9 months ago

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I suppose this is problem solving calculus. You don't regularly learn stuff this hard in school.

Daniel Liu - 4 years, 9 months ago

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@Daniel Liu Hey are you actually 14 year old...??

Rajsuryan Singh - 4 years, 9 months ago

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