What is the integral from-2 to 2 of (sin(x))/(1+x^2) ?

The integral from-2 to 2 of (sin(x))/(1+x^2) is 0

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integral from-2 to 2 of (sin(x))/(1+x^2)

\int_{- 2}^{2} \frac{sin ( x )}{1 + x ^{2}} d x

0

Solution steps

\int_{- 2}^{2} \frac{sin ( x )}{1 + x ^{2}} d x

If f(x) is an odd function and continuous at:

[- a, a]

then

\int_{- a}^{a} f (x) d x = 0

Parity of

\frac{sin ( x )}{1 + x ^{2}} :

Odd

= 0

\int_{0}^{π} ∣ cos (x + y) ∣ d x

\int_{0}^{2} \frac{- 1}{x ^{2} + 1} d x

\int_{- 2}^{2} \frac{x ^{2}}{2} d x

\int_{1}^{2} (x - 2) d x

\int_{- 5}^{8} (10 - 5 x) d x

What is the integral from-2 to 2 of (sin(x))/(1+x^2) ?
The integral from-2 to 2 of (sin(x))/(1+x^2) is 0