Integration Grandmaster - P11

$\int_0^1 \frac{1}{\sin^6 x + \cos^6 x} dx= \int_0^1 \frac{1}{1-3\sin^2 x\cos^2 x}dx$ Taking $2x= u$ , $=\int_0^1 \frac{4}{4-3\sin^2(2x)} dx=\int_0^2 \frac{2}{4-3\sin^2 u}du$ The integral is same as $\int_0^2\frac{2}{4\cos^2 u+\sin^2 u} du=2\int_0^2 \frac{\cosec^2 u}{4\cot^2u +1} du$ taking $cotu=t$ the integral becomes ,

$=-\frac{1}{2} \int_∞^{\cot(2)} \frac{dt}{t^2 +\frac{1}{4}}$ $= [-\tan^{-1}(2t)]_∞^{\cot2}$ $\implies{\boxed{\frac{π}{2} -\tan^{-1}(2\cot2)}}$ Answer is $\color{#20A900}[10000×(\frac{π}{2} -\tan^{-1}(2\cot2))]=23120$

$\begin{aligned} A & = \int_0^1 \frac 1\blue{\sin^6 x + \cos^6 x} dx & \small \blue{\text{Note that }a^3+b^3 = (a+b)(a^2+b^2-ab)} \\ & = \int_0^1 \frac 1{(\sin^2 x + \cos^2 x)\left((\sin^2 x + \cos^2 x)^2 - 3\sin^2 x \cos^2 x\right)} dx & \small \blue{= (a+b)((a+b)^2 - 2ab)} \\ & = \int_0^1 \frac 1{1-3\sin^2 x \cos^2 x} dx & \small \blue{\text{Multiply up and down by }\sec^4 x.} \\ & = \int_0^1 \frac {\sec^4 x}{\sec^4 x - 3 \tan^2 x} dx = \int_0^1 \frac {(1+\tan^2 x)\sec^2 x}{(1+\tan^2 x)^2 - 3 \tan^2 x} dx & \small \blue{\text{Let }t = \tan x \implies dt = \sec^2 x \ dx} \\ & = \int_0^{\tan 1} \frac {1+t^2}{(1+t^2)^2 -3t^2} dt \\ & = \int_0^{\tan 1} \frac {t^2+1}{(t^2-\sqrt 3t + 1)(t^2+\sqrt 3t+1)} dt & \small \blue{\text{By partial fraction decomposition}} \\ & = \int_0^{\tan 1} \frac 12 \left(\frac 1{t^2-\sqrt 3t + 1} + \frac 1{t^2+\sqrt 3t+1} \right) dt \\ & = \int_0^{\tan 1} \frac 12 \left(\frac 1{\left(t- \frac {\sqrt 3}2 \right)^2 + \frac 14} + \frac 1{\left(t+\frac {\sqrt 3}2 \right)^2 + \frac 14} \right) dt \\ & = \int_0^{\tan 1} \left(\frac 2{(2t- \sqrt 3)^2 + 1} + \frac 2{(2t+\sqrt 3)^2 + 1} \right) dt \\ & = \tan^{-1}(2t - \sqrt 3) + \tan^{-1} (2t + \sqrt 3) \ \bigg|_0^{\tan 1} \\ & = \tan^{-1} (2\tan 1 - \sqrt 3) + \frac \pi 3 + \tan^{-1} (2\tan 1 + \sqrt 3) - \frac \pi 3 \\ & = \tan^{-1} (\sqrt 3 + 2\tan 1) - \tan^{-1} (\sqrt 3-2\tan 1) \\ & \approx 2.31200 \end{aligned}$

Therefore $\lfloor 10000A\rfloor = \boxed{23120}$ .