+9519 Today our 'gifted maths' class teacher gave us a hard integration problem: \(\displaystyle\int \sqrt{\tan x}\mathtt{dx}\). Below is my solution to the question. Someone please check the answer for me? :)
\(\displaystyle\int \sqrt{\tan x}\mathtt{dx}\\ \text{Let u =}\sqrt{\tan x}\\ u^2 = \tan x\\ 2u\dfrac{\mathtt{du}}{\mathtt{dx}}=\sec^2 x\\ \mathtt{dx} = \dfrac{2u\mathtt{du}}{\sec^2 x}\\ \dfrac{\text{opposite}}{\text{adjacent}}=\tan x = u^2\\ \text{opposite} = u^2\text{, }\text{adjacent} = 1\\ \therefore \text{hypotenuse} = \sqrt{1+u^4}\\ \sec^2 x = \dfrac{\text{hypotenuse}^2}{\text{adjacent}^2}=1+u^4\)
\(\displaystyle\int \sqrt{\tan x}\mathtt{dx}\\ =2\displaystyle\int\dfrac{u^2\mathtt{du}}{\sec^2 x}\\ =2\displaystyle\int\dfrac{u^2\mathtt{du}}{u^4+1} \\ =2\displaystyle\int\dfrac{\mathtt{du}}{u^2 + \frac{1}{u^2}} \\=\displaystyle\int\dfrac{2\mathtt{du}}{u^2 + \frac{1}{u^2}}\\ =\displaystyle\int\dfrac{1+1}{u^2 + \frac{1}{u^2}}\mathtt{du}\\ =\displaystyle\int\dfrac{\left(1+\frac{1}{u^2}\right)+\left(1-\frac{1}{u^2}\right)}{u^2 + \frac{1}{u^2}}\mathtt{du}\\ =\displaystyle\int\left(\dfrac{1+\frac{1}{u^2}}{u^2+\frac{1}{u^2}}+\dfrac{1-\frac{1}{u^2}}{u^2+\frac{1}{u^2}}\right)\mathtt{du} \)
\(\quad\text{Let t = }u-\dfrac{1}{u}\\ \quad\dfrac{\mathtt{dt}}{\mathtt{du}}=1+\dfrac{1}{u^2}\\ \quad\mathtt{dt} = \left(1+\dfrac{1}{u^2}\right)\mathtt{du}\\ =\displaystyle\int\dfrac{\mathtt{dt}}{t^2+2}+\displaystyle\int\dfrac{1+\frac{1}{u^2}-\frac{2}{u^2}}{u^2+\frac{1}{u^2}}\mathtt{du}\\ \quad\text{Let } t = \sqrt2\tan v\\ \quad v = \arctan \left(\dfrac{t}{\sqrt2}\right)\\ \quad \mathtt{dt}=\sqrt2 \sec^2 v \mathtt{dv}\\ =\displaystyle\int\dfrac{\sqrt2\sec^2 v\mathtt{dv}}{2\sec^2 v}+\displaystyle\int\dfrac{1+\frac{1}{u^2}}{u^2+\frac{1}{u^2}}\mathtt{du}-\displaystyle\int\dfrac{\frac{2}{u^2}}{u^2+\frac{1}{u^2}}\mathtt{du}\\ =\dfrac{1}{\sqrt2}\arctan\left(\dfrac{t}{\sqrt2}\right)+\displaystyle\int\dfrac{t}{t^2+2}\mathtt{dt}-\displaystyle\int\dfrac{\frac{2}{u^2}}{u^2+\frac{1}{u^2}}\mathtt{du}\\ =\dfrac{1}{\sqrt2}\arctan\left(\dfrac{t}{\sqrt2}\right)+\dfrac{\ln|t^2+2|}{2}-2\displaystyle\int\dfrac{1}{u^4 +1}\mathtt{du}\\ =\dfrac{1}{\sqrt2}\arctan\left(\dfrac{u^2-1}{\sqrt2u}\right)+\dfrac{\ln|u^2+\frac{1}{u^2}|}{2}-2\displaystyle\int\dfrac{1}{u^4+1}\)
\(\quad\text{Let }j = u-\dfrac{1}{u}\\ \quad \text{Let }k = u+\dfrac{1}{u}\\ \quad \mathtt{dj} = (1+\dfrac{1}{u^2})\mathtt{du}\\ \quad \mathtt{dk} = (1 - \dfrac{1}{u^2})\mathtt{du}\\ =\dfrac{1}{\sqrt2}\arctan\left(\dfrac{u^2-1}{\sqrt2u}\right)+\dfrac{\ln|u^2+\frac{1}{u^2}|}{2}-\displaystyle\int\dfrac{u^2+1-u^2+1}{u^4+1}\mathtt{du}\\ =\dfrac{1}{\sqrt2}\arctan\left(\dfrac{u^2-1}{\sqrt2u}\right)+\dfrac{\ln|u^2+\frac{1}{u^2}|}{2}-\displaystyle\int\dfrac{u^2+1}{u^4+1}\mathtt{du}+\displaystyle\int\dfrac{u^2-1}{u^4+1}\mathtt{du}\\ =\dfrac{1}{\sqrt2}\arctan\left(\dfrac{u^2-1}{\sqrt2u}\right)+\dfrac{\ln|u^2+\frac{1}{u^2}|}{2}-\displaystyle\int\dfrac{1+\frac{1}{u^2}}{u^2+\frac{1}{u^2}}\mathtt{du}+\displaystyle\int\dfrac{1-\frac{1}{u^2}}{u^2+\frac{1}{u^2}}\mathtt{du}\\ =\dfrac{1}{\sqrt2}\arctan\left(\dfrac{u^2-1}{\sqrt2u}\right)+\dfrac{\ln|u^2+\frac{1}{u^2}|}{2}-\displaystyle\int\dfrac{\mathtt{dj}}{j^2+2}+\displaystyle\int\dfrac{\mathtt{dk}}{k^2+2}\\ =\dfrac{1}{\sqrt2}\arctan\left(\dfrac{u^2-1}{\sqrt2u}\right)+\dfrac{\ln|u^2+\frac{1}{u^2}|}{2}-\dfrac{\ln|u^2+\frac{1}{u^2}-2|}{2}+\dfrac{\ln|u^2+\frac{1}{u^2}+2|}{2}+C\\ =\dfrac{1}{\sqrt2}\arctan\left(\dfrac{\tan x-1}{\sqrt{2\tan x}}\right)+\dfrac{\ln|\tan x+\cot x|}{2}-\dfrac{\\tan x+\cot x-2|}{2}+\dfrac{\ln|\tan x+\cot x+2|}{2}+C\) \(\therefore \displaystyle\int \sqrt{\tan x}=\dfrac{1}{\sqrt2}\arctan\left(\dfrac{\tan x-1}{\sqrt{2\tan x}}\right)+\dfrac{\ln|\tan x+\cot x|}{2}-\dfrac{\\tan x+\cot x-2|}{2}+\dfrac{\ln|\tan x+\cot x+2|}{2}+C\)
+36916 I would recommend (as I mentioned a couple of days ago) checking out integral-calculator.com.....
+36916 I would recommend (as I mentioned a couple of days ago) checking out integral-calculator.com.....
SORRY MAX! EVEN MUCH LONGER THAN YOURS.
Take the integral:
integral sqrt(tan(x)) dx
For the integrand sqrt(tan(x)), substitute u = tan(x) and du = sec^2(x) dx:
= integral sqrt(u)/(u^2 + 1) du
For the integrand sqrt(u)/(u^2 + 1), substitute s = sqrt(u) and ds = 1/(2 sqrt(u)) du:
= 2 integral s^2/(s^4 + 1) ds
For the integrand s^2/(s^4 + 1), use partial fractions:
= 2 integral(-s/(2 sqrt(2) (-s^2 + sqrt(2) s - 1)) - s/(2 sqrt(2) (s^2 + sqrt(2) s + 1))) ds
Integrate the sum term by term and factor out constants:
= -1/sqrt(2) integral s/(s^2 + sqrt(2) s + 1) ds - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
Rewrite the integrand s/(s^2 + sqrt(2) s + 1) as (2 s + sqrt(2))/(2 (s^2 + sqrt(2) s + 1)) - 1/(sqrt(2) (s^2 + sqrt(2) s + 1)):
= -1/sqrt(2) integral((2 s + sqrt(2))/(2 (s^2 + sqrt(2) s + 1)) - 1/(sqrt(2) (s^2 + sqrt(2) s + 1))) ds - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
Integrate the sum term by term and factor out constants:
= -1/(2 sqrt(2)) integral(2 s + sqrt(2))/(s^2 + sqrt(2) s + 1) ds + 1/2 integral1/(s^2 + sqrt(2) s + 1) ds - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
For the integrand (2 s + sqrt(2))/(s^2 + sqrt(2) s + 1), substitute p = s^2 + sqrt(2) s + 1 and dp = (2 s + sqrt(2)) ds:
= -1/(2 sqrt(2)) integral1/p dp + 1/2 integral1/(s^2 + sqrt(2) s + 1) ds - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
The integral of 1/p is log(p):
= -(log(p))/(2 sqrt(2)) + 1/2 integral1/(s^2 + sqrt(2) s + 1) ds - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
For the integrand 1/(s^2 + sqrt(2) s + 1), complete the square:
= -(log(p))/(2 sqrt(2)) + 1/2 integral1/((s + 1/sqrt(2))^2 + 1/2) ds - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
For the integrand 1/((s + 1/sqrt(2))^2 + 1/2), substitute w = s + 1/sqrt(2) and dw = ds:
= -(log(p))/(2 sqrt(2)) + 1/2 integral1/(w^2 + 1/2) dw - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
Factor 1/2 from the denominator:
= -(log(p))/(2 sqrt(2)) + 1/2 integral2/(2 w^2 + 1) dw - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
Factor out constants:
= -(log(p))/(2 sqrt(2)) + integral1/(2 w^2 + 1) dw - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
For the integrand 1/(2 w^2 + 1), substitute v = sqrt(2) w and dv = sqrt(2) dw:
= -(log(p))/(2 sqrt(2)) + 1/sqrt(2) integral1/(v^2 + 1) dv - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
The integral of 1/(v^2 + 1) is tan^(-1)(v):
= (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
Rewrite the integrand s/(-s^2 + sqrt(2) s - 1) as 1/(sqrt(2) (-s^2 + sqrt(2) s - 1)) - (sqrt(2) - 2 s)/(2 (-s^2 + sqrt(2) s - 1)):
= (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) - 1/sqrt(2) integral(1/(sqrt(2) (-s^2 + sqrt(2) s - 1)) - (sqrt(2) - 2 s)/(2 (-s^2 + sqrt(2) s - 1))) ds
Integrate the sum term by term and factor out constants:
= (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + 1/(2 sqrt(2)) integral(sqrt(2) - 2 s)/(-s^2 + sqrt(2) s - 1) ds - 1/2 integral1/(-s^2 + sqrt(2) s - 1) ds
For the integrand (sqrt(2) - 2 s)/(-s^2 + sqrt(2) s - 1), substitute z_1 = -s^2 + sqrt(2) s - 1 and dz_1 = (sqrt(2) - 2 s) ds:
= (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + 1/(2 sqrt(2)) integral1/z_1 dz_1 - 1/2 integral1/(-s^2 + sqrt(2) s - 1) ds
The integral of 1/z_1 is log(z_1):
= (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) - 1/2 integral1/(-s^2 + sqrt(2) s - 1) ds
For the integrand 1/(-s^2 + sqrt(2) s - 1), complete the square:
= (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) - 1/2 integral1/(-(s - 1/sqrt(2))^2 - 1/2) ds
For the integrand 1/(-(s - 1/sqrt(2))^2 - 1/2), substitute z_2 = s - 1/sqrt(2) and dz_2 = ds:
= (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) - 1/2 integral1/(-z_2^2 - 1/2) dz_2
Factor -1/2 from the denominator:
= (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) - 1/2 integral2/(-2 z_2^2 - 1) dz_2
Factor out constants:
= (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) - integral1/(-2 z_2^2 - 1) dz_2
Factor -1 from the denominator:
= (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) + integral1/(2 z_2^2 + 1) dz_2
For the integrand 1/(2 z_2^2 + 1), substitute z_3 = sqrt(2) z_2 and dz_3 = sqrt(2) dz_2:
= (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) + 1/sqrt(2) integral1/(z_3^2 + 1) dz_3
The integral of 1/(z_3^2 + 1) is tan^(-1)(z_3):
= -(log(p))/(2 sqrt(2)) + (tan^(-1)(v))/sqrt(2) + (log(z_1))/(2 sqrt(2)) + (tan^(-1)(z_3))/sqrt(2) + constant
Substitute back for z_3 = sqrt(2) z_2:
= -(log(p))/(2 sqrt(2)) + (tan^(-1)(v))/sqrt(2) + (log(z_1))/(2 sqrt(2)) + (tan^(-1)(sqrt(2) z_2))/sqrt(2) + constant
Substitute back for z_2 = s - 1/sqrt(2):
= -(log(p))/(2 sqrt(2)) - (tan^(-1)(1 - sqrt(2) s))/sqrt(2) + (tan^(-1)(v))/sqrt(2) + (log(z_1))/(2 sqrt(2)) + constant
Substitute back for z_1 = -s^2 + sqrt(2) s - 1:
= -(log(p))/(2 sqrt(2)) + (log(-s^2 + sqrt(2) s - 1))/(2 sqrt(2)) - (tan^(-1)(1 - sqrt(2) s))/sqrt(2) + (tan^(-1)(v))/sqrt(2) + constant
Substitute back for v = sqrt(2) w:
= -(log(p))/(2 sqrt(2)) + (log(-s^2 + sqrt(2) s - 1))/(2 sqrt(2)) - (tan^(-1)(1 - sqrt(2) s))/sqrt(2) + (tan^(-1)(sqrt(2) w))/sqrt(2) + constant
Substitute back for w = s + 1/sqrt(2):
= -(log(p))/(2 sqrt(2)) + (log(-s^2 + sqrt(2) s - 1))/(2 sqrt(2)) - (tan^(-1)(1 - sqrt(2) s))/sqrt(2) + (tan^(-1)(sqrt(2) s + 1))/sqrt(2) + constant
Substitute back for p = s^2 + sqrt(2) s + 1:
= (log(-s^2 + sqrt(2) s - 1))/(2 sqrt(2)) - (log(s^2 + sqrt(2) s + 1))/(2 sqrt(2)) - (tan^(-1)(1 - sqrt(2) s))/sqrt(2) + (tan^(-1)(sqrt(2) s + 1))/sqrt(2) + constant
Substitute back for s = sqrt(u):
= (log(-u + sqrt(2) sqrt(u) - 1))/(2 sqrt(2)) - (log(u + sqrt(2) sqrt(u) + 1))/(2 sqrt(2)) - (tan^(-1)(1 - sqrt(2) sqrt(u)))/sqrt(2) + (tan^(-1)(sqrt(2) sqrt(u) + 1))/sqrt(2) + constant
Substitute back for u = tan(x):
= -(tan^(-1)(1 - sqrt(2) sqrt(tan(x))))/sqrt(2) + (tan^(-1)(sqrt(2) sqrt(tan(x)) + 1))/sqrt(2) + (log(-tan(x) + sqrt(2) sqrt(tan(x)) - 1))/(2 sqrt(2)) - (log(tan(x) + sqrt(2) sqrt(tan(x)) + 1))/(2 sqrt(2)) + constant
Factor the answer a different way:
= (-2 tan^(-1)(1 - sqrt(2) sqrt(tan(x))) + 2 tan^(-1)(sqrt(2) sqrt(tan(x)) + 1) + log(-tan(x) + sqrt(2) sqrt(tan(x)) - 1) - log(tan(x) + sqrt(2) sqrt(tan(x)) + 1))/(2 sqrt(2)) + constant
An alternative form of the integral is:
= (-2 tan^(-1)(1 - sqrt(2) sqrt(tan(x))) + 2 tan^(-1)(sqrt(2) sqrt(tan(x)) + 1) + log(-(tan(x) - sqrt(2) sqrt(tan(x)) + 1)/(tan(x) + sqrt(2) sqrt(tan(x)) + 1)))/(2 sqrt(2)) + constant
Which is equivalent for restricted x values to:
Answer: |= (-2 tan^(-1)(1 - sqrt(2) sqrt(tan(x))) + 2 tan^(-1)(sqrt(2) sqrt(tan(x)) + 1) + log(tan(x) - sqrt(2) sqrt(tan(x)) + 1) - log(tan(x) + sqrt(2) sqrt(tan(x)) + 1))/(2 sqrt(2)) + constant
+118609 Max, do you have access to Wolfram|Alpha,
the pay version?
I did it there.
I've downloaded the answer to my computer but Tiny Pics refuses to upload it - probably becasue it is very long .
So I can't post it easily.
Do you want me to continue to try and post it here or have you got the problem sorted now.
(I have not personally worked through any of the mathematics for it.)
