#### To determine

**To evaluate:**

∫-11sinx1+x2 dx is it exists

#### Answer

0

#### Explanation

**1) Concept:**

By using theorem for integration of an odd function

**2) Theorem:**

Suppose f is continuous on [-a, a] and if f is odd [f-x=-fx], then

∫-aaf(x)dx=0

**3) Given:**

∫-11sinx1+x2 dx

**4) Calculation:**

Consider,

∫-11sinx1+x2 dx

Here,

fx=sinx1+x2

Now,

f-x=sin(-x)1+(-x)2

Since sin-x=-sinx

f-x=-sinx1+-x2=-f(x)

Therefore, the function is odd; also it is continuous in thegiven interval, therefore, its integral exists.

From the theorem

∫-11sinx1+x2 dx=0

**Conclusion:**

Therefore,

∫-11sinx1+x2 dx=0