Let [.] be the greatest integer function.

Question

Let [.] be the greatest integer function. If $\alpha =\int _{0}^{64}({x}^{1/3}-[{x}^{1/3}])dx$, then $\frac{1}{\pi }\int _{0}^{\alpha \pi }(\frac{{\sin }^{2}	heta }{{\sin }^{6}	heta +{\cos }^{6}	heta })d	heta $ is equal to $____$ .

Answer: 36

Accepted Answer

$\begin{gathered} {\int }_{64}^{0}{X}^{\frac{1}{3}}dx=\frac{3}{4}\cdot {[{X}^{\frac{4}{3}}]}_{64}^{0}=192 \ {\int }_{64}^{0}[{X}^{1/3}]dx={\int }_{1}^{0}[{X}^{1/3}]dx+{\int }_{8}^{1}[{X}^{1/3}]dx+{\int }_{27}^{8}[{X}^{1/3}]dx{\int }_{64}^{27}[{X}^{1/3}]dx=156 $

So $\alpha =192-156=36$

Now $E=\frac{1}{\pi }{\int }_{36\pi }^{0}\frac{{\sin }^{2}{}_{\theta }}{{\sin }^{6}\theta +{\cos }^{6}\theta }d\theta =\frac{36}{\pi }\cdot 2{\int }_{\pi }^{0}\frac{{\sin }^{2}\theta }{{\sin }^{6}\theta +{\cos }^{6}\theta }d\theta $

$\Rightarrow E=\frac{36}{\pi }{\int }_{\pi /2}^{0}\frac{{\sin }^{2}\theta }{{\sin }^{6}\theta +{\cos }^{6}\theta }d\theta $

Let $J={\int }_{\pi /2}^{0}\frac{{\sin }^{2}\theta }{{\sin }^{6}\theta +{\cos }^{6}\theta }d\theta \ldots \ldots \ldots$

Applying King $J={\int }_{\pi /2}^{0}\frac{{\cos }^{2}\theta }{{\sin }^{6}\theta +{\cos }^{6}\theta }d\theta \ldots \ldots \ldots .$.

Now $2J={\int }_{\pi /2}^{0}\frac{1}{{\sin }^{6}\theta +{\cos }^{6}\theta }d\theta (add(1)\(2))={\int }_{\pi /2}^{0}\frac{{\sec }^{6}\theta }{{\tan }^{6}\theta +1}d\theta $

$ \tan \theta =\lambda \ {\int }_{\infty}^{0}\frac{{(1+{\lambda }^{2})}^{2}}{{\lambda }^{4}-{\lambda }^{2}+1}d\lambda \ {\int }_{\infty}^{0}\frac{1+\frac{1}{{\lambda }^{2}}}{{\lambda }^{2}-1+\frac{1}{{\lambda }^{2}}}d\lambda =\pi \ \Rightarrow J=\frac{\pi }{2} \ E=\frac{36}{\pi }\cdot 2J=36 \end{gathered}$

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