Definite integral of an even function

[Samuelb88]

Homework Statement

Integrate the definite integral

$$\int_{-2}^{2}{\frac{x^2}{4+x^6} dx$$

Homework Equations

The Attempt at a Solution

(1) The integrand f is an even function, therefore:

$$2\int_{0}^{2}{\frac{x^2}{4+x^6} dx$$

(2) I re-expressed the denominator as:

$$2\int_{0}^{2}{\frac{x^2}{4+(x^3)^2} dx$$

(3) I made the t-substitution:

$$t=x^3$$

$$\frac{1}{3}\right) dt = x^2dt$$

$$\frac{2}{3}\right) \int_{0}^{8} {\frac{1}{4+t^2} dt$$

(4) Here's where I get stuck. I can seem to make another substitution to be able to simplify the integral such that it can be evaluated or be able use integration by parts to be able to evaluate it.

 

[Mentallic]

That integral can be found on any standard integrals sheet such that:

$$\int {\frac{1}{4+t^2} dt$$

$$=\frac{1}{2}tan^{-1}(\frac{t}{2})+c$$

or are you unsatisfied with this?

 

[Mark44]

Homework Statement

Integrate the definite integral

$$\int_{-2}^{2}{\frac{x^2}{4+x^6} dx$$

Homework Equations

The Attempt at a Solution

(1) The integrand f is an even function, therefore:

$$2\int_{0}^{2}{\frac{x^2}{4+x^6} dx$$

(2) I re-expressed the denominator as:

$$2\int_{0}^{2}{\frac{x^2}{4+(x^3)^2} dx$$

(3) I made the t-substitution:

$$t=x^3$$

$$\frac{1}{3}\right) dt = x^2dt$$

$$\frac{2}{3}\right) \int_{0}^{8} {\frac{1}{4+t^2} dt$$

(4) Here's where I get stuck. I can seem to make another substitution to be able to simplify the integral such that it can be evaluated or be able use integration by parts to be able to evaluate it.

For your last integral you need to to a trig substitution, or else know this integration formula (which can be derived by a trig substitution):
$$\int \frac{dx}{a^2 + x^2}~=~\frac{1}{a} tan^{-1}(x/a) + C$$

 

[Samuelb88]

ahh, thanks guys. we're going to start trig. substitution next week so i guess I'm satisfied for now. :)