Definite integral involving odd and even function.

Question

$$\int_{\frac {-1}{\sqrt 3}}^{\frac {1}{\sqrt 3}} \frac {x^4}{1-x^4}\cos^{-1}(\frac {2x}{1+x^2})dx$$

As the limits are $a$ and $-a$, I try to figure out what is $f(-x)$,but the function is neither becoming $-f(x)$ or $f(x)$. Note that ,I first converted $cos^{-1}x$ to $sin^{-1}x$.

@Andrew Li Can you please edit my post,actually i donot yet fully acquainted with jax — Sumit Gupta, Mar 17 '18 at 14:15
I would recommend getting at least acquainted with it before posting because I don't have the time currently to do so. Have you seen the basic tutorial on Meta? — Andrew Li, Mar 17 '18 at 14:17
Thanks @Leyla Alkan Could you tell me how to approach the problem. — Sumit Gupta, Mar 17 '18 at 14:21

score 1 · Answer 1 · answered Mar 17 '18 at 14:33

Like Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}dx$,

$$I=\int_{-a}^a\dfrac{x^{2n}}{1-x^{2n}}\arccos\dfrac{2x}{1+x^2}dx$$

$$=\int_{-a}^a\dfrac{x^{2n}}{1-x^{2n}}\arccos\dfrac{(-2x)}{1+x^2}dx$$

Using How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?,

$$I+I=\pi\int_{-a}^a\dfrac{x^{2n}}{1-x^{2n}}dx$$

Now $$\int\dfrac{x^{2n}}{1-x^{2n}}dx=\int\dfrac{dx}{1-x^{2n}}-\int dx$$

Definite integral involving odd and even function.

1 Answers1