Question

\(\int\limits_0^1 {\int\limits_{\arcsin y}^{{\raise0.7ex\hbox{\(\pi \)} \!\mathord{\left/ {\vphantom{\pi 2}}\right.}\!\lower0.7ex\hbox{\(2\)}}} {\cos x\sqrt {1 + {{\cos }^2}x} dxdy} } \).

Short answer

The process of switching between \(dxdy\) order and \(dydx\) order in double integral is called changing the order of integration.

Step-by-step solution

Given Data.

Consider the integral:

\(\int\limits_0^1 {\int\limits_{\arcsin y}^{{\raise0.7ex\hbox{$\pi $} \!\mathord{\left/

{\vphantom {\pi 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$2$}}} {\cos x\sqrt {1 + {{\cos }^2}x} dx} } dy\)

The inner limits of integration indicate that \(\arcsin y \le x \le \frac{\pi }{2}\) and the outer limits of integration indicate that \(0 \le y \le 1\).

Graphical Representation.

Now integrate first on \(y\), start by finding the limits of a vertical line passing through the region of integration \(0 \le y \le \sin x\).

The range of \(x\) values is \(0 \le x \le \frac{\pi }{2}\).

Call the integral I and set it up in reverse order.

Reversing Order of Integration.

\(\begin{array}{c}\int\limits_0^1 {\int\limits_{\arcsin y}^{{\raise0.7ex\hbox{$\pi $} \!\mathord{\left/

{\vphantom {\pi 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$2$}}} {\cos x\sqrt {1 + {{\cos }^2}x} dx} } dy = \int\limits_0^{{\raise0.7ex\hbox{$\pi $} \!\mathord{\left/

{\vphantom {\pi 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$2$}}} {\int\limits_0^{\sin x} {\cos x\sqrt {1 + {{\cos }^2}x} } } dydx\\ = \int\limits_0^{{\raise0.7ex\hbox{$\pi $} \!\mathord{\left/

{\vphantom {\pi 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$2$}}} {\cos x\sin x} dx\end{array}\)

Put \(u = \cos x \Rightarrow du = - \sin xdx\).

When \(x \to 0 \Rightarrow u \to 1\)

When \(x \to \frac{\pi }{2} \Rightarrow u \to 0\)

\(\begin{array}{c}\int\limits_0^1 {\int\limits_{\arcsin y}^{{\raise0.7ex\hbox{$\pi $} \!\mathord{\left/

{\vphantom {\pi 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$2$}}} {\cos x\sqrt {1 + {{\cos }^2}x} dx} } dy = \int\limits_0^1 u du\\ = \left( {\frac{{{u^2}}}{2}} \right)_0^1\\ = \frac{1}{2}\end{array}\)

Therefore, the value of integral is \(\frac{1}{2}\).