Question

for calculus enthusiasts...

\(\large \int (\sec x + \csc x)^2 dx\)

how to solve?

Answer 1

Open it/ Expand it

Answer 2

i got \(\large \cot x - 2\int \frac{1}{\sin x \cos x} + \tan x\)

Answer 3

\[4\int \frac{1}{\sin 2x}\]

Answer 4

\[4 \int \csc 2x \]

Answer 5

\(\Large \mathtt{\color{pink}{uh} \color{lightgreen}{-} \color{lightblue}{huh}}\)

Answer 6

\[4 \int\limits \csc^2 2x\quad dx+4\int\limits \csc2x \quad dx\]

Answer 7

^where did that come from?

Answer 8

i think cinar posted it, but i cant be too sure :P

Answer 9

i cant see the post @_@

Answer 10

\[(\sec(x)+\csc(x))^2=\sec^2(x)+2\sec(x)\csc(x)+\csc^2(x)\]
\[=\sec^2(x)+2\frac{1}{\cos(x)}\frac{1}{\sin(x)}+\csc^2(x)\]
\[=\sec^2(x)+\frac{2}{\cos(x) \sin(x)}+\csc^2(x)\]
\[=\sec^2(x)+\frac{2 \cdot 2}{2 \cos(x) \sin(x)} +\csc^2(x)\]
\[=\sec^2(x)+\frac{4}{\sin(2x)}+\csc^2(x)\]
\[=\sec^2(x)+4 \csc(2x)+\csc^2(x)\]

Answer 11

this looks fun im too late though :(

Answer 12

at 1245am, youre all too late

Answer 13

or is that 0045 hours?

Answer 14

its whatever you want it to be

Answer 15

0075 if we use military metric lol

Answer 16

yes @myininaya the first and last terms are very much integrable...but the middle term i dont know how :C

Answer 17

might have an error >.<

Answer 18

\[\int\limits\limits_{}^{}\csc(2x) dx=\int\limits\limits_{}^{}\csc(2x) \cdot \frac{\csc(2x)+\cos(2x)}{\csc(2x)+\cot(2x)} dx \]
Let \[u=\csc(2x)+\cot(2x) => du=-2 \csc(2x)(\cot(2x)+\csc(2x)) dx\]

Answer 19

\[-2cosec2xcos2x-2\ln|cosec2x+\cot2x|+C\]

Answer 20

hmm @cinar there's still that \(4\int \csc (2x)\) problem in your solution :p

Answer 21

\[\int\limits_{}^{}\frac{du}{-2 u}=\frac{-1}{2} \ln|u|+C\]

Answer 22

i dont get the multiplication thing you did myin :P im sure that wasnt conjugate...

Answer 23

can't you use the table..

Answer 24

for integrating sec(x) and csc(x) you need to multiply by a weird one

Answer 25

my post was only for the:\[2\int\limits\sec x \csc xdx \]term after you square.

Answer 26

for sec(x) it is (sec(x)+tan(x))/(sec(x)+tan(x))
and
for csc(x) it is (csc(x)+cot(x))/(csc(x)+cot(x))

Answer 27

That's quite neat @joemath314159. Always like tricks like that :-)

Answer 28

ah, the signs are backwards though, ln(cos) should be the negative one.

Answer 29

hmm seems @joemath314159 gave the most convincing and comprehensible reply lol..thanks :DDD

Answer 30

though i think the one multiplied to csc x is cscx - cotx @myininaya i mean that's what i remember..is it +?