Question

intergral of csc 5pix cot 5pix dx in the interval of (1/30, 1/10)

Answer 1

here's a hint... \[\large \int \csc 5\pi x \cot 5\pi x dx \implies \int \frac 1{\sin 5\pi x} \times \frac{\cos 5\pi x}{\sin 5\pi x}dx\]
\[\large \implies \int \frac{\cos 5\pi x}{\sin^2 5\pi x} dx\]does that give you any ideas?

Answer 2

\[\int\limits_{1/30}^{1/10} \csc(5\pi x)\cot(5\pi x) dx\] Let: \[u = 5\pi x\] Therefore: \[du = 5\pi dx\] Just divide out the 5pi: \[\frac{ 1 }{ 5\pi } \int\limits_{1/30}^{1/10} \csc(u)\cot(u)du = \frac{ 1 }{ 5\pi }(-\csc(u))|[1/30,1/10]\] Resubstitute in function for u: \[=\frac{ 1 }{ 5\pi }(-\csc(5\pi x))|[1/30,1/10]\] Substitute in values: \[=\frac{ 1 }{ 5\pi }[(-\csc(5\pi*1/10) - (-\csc(5\pi * 1/30)]\] Solve it down: \[=\frac{ 1 }{ 5\pi }(\frac{ -1 }{ \sin(\pi / 2)} + \frac{ 1 }{ \sin(\pi / 6) }) = \frac{ 1 }{ 5\pi }(\frac{ -1 }{ \frac{ \sqrt{2} }{ 2 } }+\frac{ 1 }{ \frac{ 1 }{ 2 } })\] Which is (almost done LOL): \[=\frac{ 1 }{ 5\pi } (\frac{ -2 }{ \sqrt{2} }+2) = \frac{ 1 }{ 5\pi } (\frac{ -4\sqrt{2} }{ 2 } + \frac{ 4 }{ 2 }) = \frac{ 1 }{ 5\pi }(\frac{ -4\sqrt{2}+4 }{ 2 })\] Finally is: \[=\frac{ -4\sqrt{2}+4 }{ 10\pi }= \frac{ -2\sqrt{2}+2 }{ 5\pi }\]
Wow.... why was that so long... hate trig xD

Answer 3

@nj1202 Let do the easy way, follow what @lgbasallote shows:
Let u = sin5πx --> du = - 5π cos 5π dx

Answer 4

That's why math is awesome! Multiple ways to a single answer :D

Answer 5

:)