The value of $\int_{-\pi/2}^{\pi/2}(x^5 + x^3\cos x)dx$ is

Given: $\int_{-\pi/2}^{\pi/2}(x^5 + x^3\cos x)dx$

For any odd function $f(x)$ where $f(-x) = -f(x)$:

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

The negative side cancels the positive side.

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Breaking the integral into two parts:

$\int_{-\pi/2}^{\pi/2}(x^5 + x^3\cos x)dx = \int_{-\pi/2}^{\pi/2}x^5 dx + \int_{-\pi/2}^{\pi/2}x^3\cos x \, dx$

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Consider $f(x) = x^5$

$f(-x) = (-x)^5$

$f(-x) = -x^5$

$f(-x) = -f(x)$

Therefore $x^5$ is an odd function.

$\int_{-\pi/2}^{\pi/2}x^5 dx = 0$

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Consider $g(x) = x^3\cos x$

$g(-x) = (-x)^3 \cdot \cos(-x)$

$g(-x) = -x^3 \cdot \cos x$

Since $\cos(-x) = \cos x$:

$g(-x) = -g(x)$

Therefore $x^3\cos x$ is an odd function.

$\int_{-\pi/2}^{\pi/2}x^3\cos x \, dx = 0$

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$\int_{-\pi/2}^{\pi/2}(x^5 + x^3\cos x)dx = 0 + 0$

$\int_{-\pi/2}^{\pi/2}(x^5 + x^3\cos x)dx = 0$