Question

Hey people, how can I integrate sqrt(tan x), can someone give me an idea?

Answer 1

\[\int\limits \sqrt{tanx}\]

Answer 2

Right?

Answer 3

I can't see your 1st reply

Answer 4

@Psymon

Answer 5

let tanx=t

Answer 6

Lol, alright, I wanna see this.

Answer 7

Also remember that:
\[\int\limits tanx =\ln |sexc| +C\]

Answer 8

*dx

Answer 9

Same thing Psymon -_-

Answer 10

I learned it both ways

Answer 11

Not used to seeing it that way :P

Answer 12

LALALA XD

Answer 13

I hope @DLS isn't writing a paragraph for this

Answer 14

Let \(\large \sqrt{tanx}=t\).
\[\large t^2=tanx\]
\[\large \sec^2x=2t \frac{dt}{dx}\]
\[\large dt = \frac{2t}{1+t^4}\]
\[\large \int\limits_{}^{} t \times \frac{2t}{1+t^4} dt\]
\[\large \int\limits_{}^{}( \frac{t^2+1}{1+t^4} + \frac{t^2-1}{1+t^4} )dt\]
Hope you can do now.

Answer 15

You're suppose to guide them, but at least you didn't do all of their work for them :3

Answer 16

Yup I just dropped the hint.
no one took efforts to work on my first hint so did a few more steps.

Answer 17

Btw Welcome to Openstudy @Yttrium :P

Answer 18

haha

Answer 19

Hey, how come it became ∫(t2+1/1+t4 + t2−1/1+t4)dt

Answer 20

\[\large 2t^2=t^2+1+t^2-1\]

Answer 21

What shall I do next? Partial fractions?

Answer 22

Just divide by t^2 in numerator and denominator,convert the numerator or denominator any one to a perfect whole square and let it be say z,everything will cancel out.

Answer 23

Okay I got it. :D