How do I evaluate $\int_{\frac{1}{6}}^{\frac{1}{2}} [csc (π t)] [cot (π t)] d t$ $int_(1/6)^(1/2)[csc(pit)][cot(pit)] dt$ ?

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How do I evaluate $\int_{\frac{1}{6}}^{\frac{1}{2}} [csc (π t)] [cot (π t)] d t$ $int_(1/6)^(1/2)[csc(pit)][cot(pit)] dt$ ?

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Answer 1 · 2015-03-14T01:56:45

For any integral, first try straightaway integration, then substitution. Do I know a function whose derivative is or involves csc time cot?

Yes, of course: $\frac{d}{d x} (csc x) = - csc x cot x$ $d/(dx)(cscx)=-cscxcotx$ .
That's not quite what we're integrating, but we can fix that with a substitution:
Let $u = π t$ $u=pit$ . That makes $d u = π d t$ $du=pi dt$ and $d t = \frac{d u}{π}$ $dt = (du)/pi$
When $x = \frac{1}{6}$ $x=1/6$ (the lower limit of integration), then $u = 4 \cdot \frac{1}{6} = \frac{π}{6}$ $u=4*1/6=pi/6$ .
When $x = \frac{1}{2}$ $x=1/2$ (the upper limit). then $u = \frac{π}{2}$ $u=pi/2$ . So, substituting, our integral becomes:

$\int_{\frac{π}{6}}^{\frac{π}{2}} csc u cot u \frac{d u}{π} = \frac{1}{π} \int_{\frac{π}{6}}^{\frac{π}{2}} csc u cot u d u$ $int_(pi/6)^(pi/2)cscucotu (du)/pi=1/piint_(pi/6)^(pi/2)cscucotu du$

$= \frac{1}{π} (- csc u)]_{\frac{π}{6}}^{\frac{π}{2}} = \frac{1}{π} (- csc (\frac{π}{2}) + csc (\frac{π}{6}))$ $=1/pi(-cscu)]_(pi/6)^(pi/2)=1/pi(-csc(pi/2)+csc(pi/6))$

$- \frac{1}{π} (- 1 + 2) = \frac{1}{π}$ $-1/pi(-1+2)=1/pi$

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How do I evaluate $\int_{\frac{1}{6}}^{\frac{1}{2}} [csc (π t)] [cot (π t)] d t$ $int_(1/6)^(1/2)[csc(pit)][cot(pit)] dt$ ?

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How do I evaluate $\int_{\frac{1}{6}}^{\frac{1}{2}} [csc (π t)] [cot (π t)] d t$ $int_(1/6)^(1/2)[csc(pit)][cot(pit)] dt$ ?