# Trigonometry Substitution (Integral)

1. Feb 23, 2013

### KDeep

1. The problem statement, all variables and given/known data
∫xtan^-1(x^2)dx

2. Relevant equations

3. The attempt at a solution

I did u = x, du = 1,
v = ? ,dv = tan^-1(x^2)dx

I do not know how to get the integral of tan^-1(x^2)

2. Feb 23, 2013

### LCKurtz

u = x can never simplify an integral. It just changes the name of the variable. Try $u=x^2$.

3. Feb 23, 2013

### KDeep

u = x^2
du = 2xdx
du/2 = xdx

∫xtan(x^2)dx

(1/2)∫tan(u)du

(1/2) ln |sec(x^2)| + C

Correct?

4. Feb 23, 2013

### LCKurtz

You don't have to ask. Differentiate it and check if it works. Did you forget an arctangent?

5. Feb 23, 2013

### KDeep

u = x^2
du = 2xdx
du/2 = xdx

∫xtan^1(x^2)dx

(1/2)∫tan^-1(u)du

I cannot figure out the integral of tan^-1(u)

6. Feb 23, 2013

### SammyS

Staff Emeritus
For the integral, $\displaystyle \ \ \int \tan^{-1}(t)\,dt\,,\$ use integration by parts, with u = tan-1(t) and dv = dt , else look it up. http://en.wikipedia.org/wiki/Inverse_trig_functions#Indefinite_integrals_of_inverse_trigonometric_functions