Question

Integral of sqrt(tanx) ?

Answer 1

I don't think you can. At least not in terms of elementary functions we know of

Answer 2

"Its not elegant... just long and a bit intuitive in middle." ~ http://math.stackexchange.com/questions/759567/interesting-calculus-problem-advice

Answer 3

huhm... interesting

Answer 4

wolfram alpha gave the answer, which seems like a tedious integral.

Answer 5

ok

Answer 6

no other way? @Zarkon @freckles?

Answer 7

I wonder if it would be easier to look at it like this:
\[\int\limits_{}^{}\frac{\sqrt{\tan(x)}}{\sec^2(x)} \sec^2(x) dx\]

Answer 8

write the bottom sec^2(x) in terms of tan^2(x) by pythagoran identity

Answer 9

you will end up with a rational function to integrate

Answer 10

after a sub

Answer 11

oooooo

Answer 12

but I don't know how to integrate the resulting rational function

Answer 13

http://www.wolframalpha.com/input/?i=integrate%282u%5E2%2F%281%2Bu%5E4%29%2Cu%29
actually it looks like we can use partial fractions

Answer 14

integrate sinx/cosx
i gues i did this integral not long ago i got ln|secx|+c

Answer 15

o pellet it sqrt(tanx) my bad lol

Answer 16

o pellet* i don't where that pellet came hhe
what does it mean anyway lol

Answer 17

\[2\int\frac{u^2}{1+u^4}~du\]
To get the right partial fractions, we could try expressing the denominator as the product of two degree-2 polynomials:
\[\begin{align*}u^4+1&=(u^2+au+1)(u^2+bu+1)\\\\
&=u^4+(a+b)u^3+(ab+2)u^2+(a+b)u+1
\end{align*}\]
which yields a system with solution \(a=\pm\sqrt2\) and \(b=\mp\sqrt2\). (One is positive, the other is negative.)

Curious to see if this sort of approach to decomposition would work if we assume the original polynomial can be factored into a deg-1 and deg-3 polynomial, and if that would alter the result in any significant way...

Answer 18

Let's see: Assume
\[u^4+1=(u+1)(u^3+au^2+bu+1)\]
Then
\[\begin{align*}u^4+1&=(u+1)(u^3+au^2+bu+1)\\\\
&=u^4+(1+a)u^3+(a+b)u^2+(1+b)u+1
\end{align*}\]
which yields no solution... Hmm, looks like it goes to show that PF expressions are unique.