Mathematics & Numerical Analysis

Evaluation of iterated double integral:
First evaluate the inner integral with respect to $x$:
$\int_0^{\cos y} x \sin y \, dx = \sin y \left[ \frac{x^2}{2} \right]_0^{\cos y} = \frac{1}{2} \sin y \cos^2 y$.
Now evaluate the outer integral with respect to $y$:
$\int_0^\pi \frac{1}{2} \cos^2 y \sin y \, dy$.
Let $u = \cos y \implies du = -\sin y \, dy$.
When $y = 0, u = 1$; when $y = \pi, u = -1$.
$\int_1^{-1} \frac{1}{2} u^2 (-du) = \int_{-1}^1 \frac{1}{2} u^2 du = \frac{1}{2} \left[ \frac{u^3}{3} \right]_{-1}^1 = \frac{1}{6} (1 - (-1)) = \frac{2}{6} = \frac{1}{3}$.

Option breakdown:

• $\frac{1}{2}$ is an incorrect evaluation.
• $\frac{1}{3}$ is the exact value of the double integral.
• $\frac{1}{5}$ is incorrect.
• $-\frac{1}{3}$ has an incorrect negative sign.

Euler's Beta and Gamma function relation for trigonometric integrals:
The standard formula for definite integrals involving sines and cosines is:
$\int_0^{\pi/2} \sin^p x \cos^q x \, dx = \frac{\Gamma\left(\frac{p+1}{2}\right) \Gamma\left(\frac{q+1}{2}\right)}{2 \Gamma\left(\frac{p+q+2}{2}\right)}$.
Setting $p = 0$ and $q = n$:
$\int_0^{\pi/2} \cos^n x \, dx = \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{n+1}{2}\right)}{2 \Gamma\left(\frac{n+2}{2}\right)}$.
Since $\Gamma(1/2) = \sqrt{\pi}$:
$\int_0^{\pi/2} \cos^n x \, dx = \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)} \frac{\sqrt{\pi}}{2}$.

Option breakdown:

• $\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n+2}{2})} \frac{\sqrt{\pi}}{2}$ is the standard reduction formula in terms of the Gamma function.
• Inverts the ratio of gamma arguments.
• Has argument $n/2$ instead of $(n+1)/2$ in the numerator.
• Has argument $n/2$ in the denominator instead of $(n+2)/2$.