Trigonometry Substitution (Integral)

[KDeep]

Homework Statement

∫xtan^-1(x^2)dx

Homework Equations

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)

 

[LCKurtz]

Homework Statement

∫xtan^-1(x^2)dx

Homework Equations

The Attempt at a Solution

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

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

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

 

[KDeep]

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

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?

 

[LCKurtz]

Correct?

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

 

[KDeep]

Did you forget an arctangent?

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)

 

[SammyS]

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)

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
Added in Edit:

An alternate way to approach this problem (It is titled, Trigonometry Substitution (Integral) .) is to use the substitution, x² = tan(t) .

It will also require the use of integration by parts.