What is \(\mathop \smallint \limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} x\sin x\;dx\) equal to?

Concept:

Odd and even function:

If a function f(x) such that f(-x) = f(x) then f(x) is an even function and if f(-x) = -f(x) then f(x) is an odd function.

Integration of odd and even function:

The following two cases hold for a differentiable function f(x).

If f(x) is an even function then \( \rm \displaystyle\int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx\).

If f(x) is an odd function then \(\rm \displaystyle\int_{-a}^{a}f(x)dx = 0\).

Calculation:

Let \(\rm f(x) = x\sin x\), put -x instead of x then we get \(\rm f(-x) = (-x)\sin(-x) = x\sin x\).

Therefore, f(-x) = f(x) implies that f(x) is an even function.

Thus, the integral is given by,

\(\rm \begin{align*} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x\sin x dx &= 2\int_{0}^{\frac{\pi}{2}}x\sin x dx\\ &= 2\left[x(-\cos x) - \int(-\cos xdx)\right]_{0}^{\frac{\pi}{2}}\\ &= 2\left[-x\cos x + \sin x\right]_{0}^{\frac{\pi}{2}}\\ &= 2\left[\left(-\dfrac{\pi}{2}\cos\left(\dfrac{\pi}{2}\right) + \sin\left(\dfrac{\pi}{2}\right)\right)-\left(-0\cos 0+\sin 0\right)\right]\\ &= 2 \end{align*}\)

Hence, the value of the given integral is 2.