Question

Integrate

Answer 1

\[\int\limits \sqrt{\tan x} \text{ dx}\] @Vocaloid

Answer 2

i'm completely stumped :(

Answer 3

Write tanx in form of sin and cos and then substitute \(\sqrt{cosx}=t\) and proceed

Answer 4

I'm just here to say yolo

Answer 5

I'm thinking of a shorter way to do this

Answer 6

\(\huge \int \frac{\sqrt[4]{1-t^4}}{t}\text{ dx}\)
it's not working :(

Answer 7

it's a little tedious but I believe you can start by doing a u-substitution where u = sqrt[tan(x)] therefore making u^2 = tan(x)

so we'd want to re-write the expression in terms of u only, so we need to convert the dx to a du

2u du = sec^2(x)dx [taking the deriv. of both sides]

using the tan^2 and sec^2 identity

2u du = 1 + tan^2(x)dx

since we have tan(x) again we can now re-write completely in terms of u

2u du = 1 + (u^2)^2 dx

dx = 2u du / (1 + [u^4)

so now the entire integral becomes

integral of 2u * u du / (1 + u^4)

(sorry for the sloppy notation)

Answer 8

interesting problem

Answer 9

wow you actually tried to do it lol. i'll give it to you, you're on the right track, but this is the first of at least 5 more substitutions you have to do

Answer 10

I would much rather be doing this than a million rate of change problems, believe me

Answer 11

are you sure about that?

Answer 12

I think either mimi_x3 or lgbasallote claimed to have a shorter route, but as we both know both users are long gone :(

Answer 13

I am getting \(\huge \int\limits_{}^{}\frac{- 2dt }{\sqrt[4]{1-t^4 }}\) after that substitution

Answer 14

vocaloid would you prefer a troll geometry question

Answer 15

\(tanx=t^2\) and this problem is over

Answer 16

qwerty what's your final answer

Answer 17

Wait a min

Answer 18

That's what voca did..

Answer 19

disclaimer: not my idea but you can split

integral of 2u^2 du / (1 + u^4) into [ (u^2+1) + (u^2 -1) ] / (1+u^4)

then

[ (u^2+1) ] / (1 + u^4)* + [(u^2 -1)] / (1+u^4)**

then taking the left side * and dividing by u^2 gives:

[ ( 1 + 1/u^2) ] / (1/u^2 + u^2) ]

then apparently you can do another u-substition (they use z) where dz = 1 + 1/u^2 du and z = u - 1/u

after a bit of messy algebra/substitution you get

1/z^2 + u = 1/sqrt(2) arctan(z/sqrt(2))

plug u back in for z and get

1/z^2 + u = 1/sqrt(2) arctan(/sqrt(2))

Answer 20

ugh the bb code messed up that last line

Answer 21

hmm yeah i think you needa eat this geometry problem ;)

Answer 22

then you would repeat this process with the other half of the integral

and get something like this

Answer 23

anyway if you have another one I'd love to take a look

Answer 24

well this one is basic geometry, if that floats your boat. fiendishly difficult for "basic geometry" but it's just 9th grade level geometry if you wanna take a stab at it