+0

# Integration

0
946
2
+7705

$$\displaystyle\int\sqrt{\tan x}\;\text{d}x$$

Jun 6, 2017

#1
0

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

Jun 6, 2017

#1
0

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

Guest Jun 6, 2017
#2
+468
+2

 you are right guest!Thanks Max!You even solved my problem!
Jun 6, 2017