$$\int_0^{\pi}\lfloor\pi^2\cos^3x\rfloor\sin x\,dx$$ (where $\lfloor x \rfloor $ is the floor of $x$)
I thought of breaking into required bounds but its too lengthy. Moreover I had to take cube root and then $\cos$ inverse. Please give a hint.
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user354545
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@Landuros Your edit has affected the relevance of an already-posted answer. – gen-ℤ ready to perish Nov 25 '17 at 06:28
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See https://math.stackexchange.com/questions/2523420/show-that-int-0-pi-f-sin-x-mathrmdx-2-int-0-pi-2f-sin-x-math/2523449#2523449 – lab bhattacharjee Nov 25 '17 at 07:07
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@ChaseRyanTaylor Sorry if I disrupted anything, I'm just making sure people are using correct notation. – Landuros Nov 25 '17 at 09:10
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Since,
$$ \int_0^{2a}f(x)\,dx =\int_0^{a}\bigl(f(a+x)+f(a-x)\bigr)\,dx$$
$$ \implies\int_0^{\pi}\left\lfloor\pi^2\cos^3x\right\rfloor\sin x\,dx=\int_0^{{\pi}/{2}}\left(\left\lfloor-\pi^2\sin^3x\right\rfloor+\left\lfloor\pi^2\sin^3x\right\rfloor\right)\cos x\,dx$$
Clearly, most of the terms will get cancelled out.
gen-ℤ ready to perish
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Your IDE
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That is absolutely incredible! Where did you learn that identity? – gen-ℤ ready to perish Nov 25 '17 at 06:25
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1@ChaseRyanTaylor Thank you. It was in my 12th class Mathematics and this identity itself is the foundation for
$ \int_{-a}^{a}f(x),dx = \begin{cases} 2 \int_0^{a}f(x),dx & \text{if $f$ is even} \ 0 & \text{if $f$ is odd} \end{cases} $
– Your IDE Nov 25 '17 at 06:37 -
Given @Landuros’s edit to the original question, you might wish to update your notation to incorporate
\lfloorand\rfloor– gen-ℤ ready to perish Nov 25 '17 at 06:48 -
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