(2.5) evaluate the following iterated integral: $\\int_{0}^{1}\\int_{\\arcsin y}^{\\frac{\\pi}{2}}\\cos…

Explanation:

Step1: Integrate with respect to $x$ first

Let $u = \cos x$, then $du=-\sin xdx$. When $x = \arcsin y$, $\cos x=\sqrt{1 - y^{2}}$; when $x=\frac{\pi}{2}$, $\cos x = 0$. The inner - integral $\int_{\arcsin y}^{\frac{\pi}{2}}\cos x\sqrt{1+\cos^{2}x}dx=-\int_{\sqrt{1 - y^{2}}}^{0}\sqrt{1 + u^{2}}du=\int_{0}^{\sqrt{1 - y^{2}}}\sqrt{1 + u^{2}}du$. We know that the integral formula for $\int\sqrt{a^{2}+x^{2}}dx=\frac{x}{2}\sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2}\ln(x+\sqrt{a^{2}+x^{2}})+C$ (here $a = 1$), so $\int_{0}^{\sqrt{1 - y^{2}}}\sqrt{1 + u^{2}}du=\left[\frac{u}{2}\sqrt{1 + u^{2}}+\frac{1}{2}\ln(u+\sqrt{1 + u^{2}})\right]_{0}^{\sqrt{1 - y^{2}}}=\frac{\sqrt{1 - y^{2}}}{2}\sqrt{2 - y^{2}}+\frac{1}{2}\ln(\sqrt{1 - y^{2}}+\sqrt{2 - y^{2}})$.

Step2: Integrate the result with respect to $y$

The original double - integral becomes $\int_{0}^{1}\left(\frac{\sqrt{1 - y^{2}}}{2}\sqrt{2 - y^{2}}+\frac{1}{2}\ln(\sqrt{1 - y^{2}}+\sqrt{2 - y^{2}})\right)dy$. We can also use another approach. Change the order of integration. The region of integration is given by $0\leq y\leq1,\arcsin y\leq x\leq\frac{\pi}{2}$, which can be rewritten as $0\leq x\leq\frac{\pi}{2},0\leq y\leq\sin x$. The double - integral $\int_{0}^{1}\int_{\arcsin y}^{\frac{\pi}{2}}\cos x\sqrt{1+\cos^{2}x}dxdy=\int_{0}^{\frac{\pi}{2}}\int_{0}^{\sin x}\cos x\sqrt{1+\cos^{2}x}dydx$. First, integrate with respect to $y$: $\int_{0}^{\frac{\pi}{2}}\cos x\sqrt{1+\cos^{2}x}\left[y\right]{0}^{\sin x}dx=\int{0}^{\frac{\pi}{2}}\sin x\cos x\sqrt{1+\cos^{2}x}dx$. Let $t = 1+\cos^{2}x$, then $dt=-2\sin x\cos xdx$. When $x = 0$, $t = 2$; when $x=\frac{\pi}{2}$, $t = 1$. $\int_{0}^{\frac{\pi}{2}}\sin x\cos x\sqrt{1+\cos^{2}x}dx=-\frac{1}{2}\int_{2}^{1}\sqrt{t}dt=\frac{1}{2}\int_{1}^{2}t^{\frac{1}{2}}dt$.

Step3: Calculate the final integral

Using the power - rule for integration $\int t^{n}dt=\frac{t^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\frac{1}{2}\int_{1}^{2}t^{\frac{1}{2}}dt=\frac{1}{2}\times\left[\frac{2}{3}t^{\frac{3}{2}}\right]_{1}^{2}=\frac{1}{3}(2\sqrt{2}-1)$.

Answer:

$\frac{1}{3}(2\sqrt{2}-1)$