Problem 0002
Prove that [tex]I = \int\limits_0^{\pi /2} {{{\sqrt {\sec x} } \over {\sqrt {\cos ecx }+ \sqrt {\sec x} }}} dx = {\pi \over 4}[/tex]
Answer:
[tex]I = \int\limits_0^{\pi /2} {{{\sqrt {\sec x} } \over {\sqrt {\cos ec x }+ \sqrt {\sec x} }}} dx[/tex]....... (1)
or [tex]I = \int\limits_0^{\pi /2} {{{\sqrt {\sec (\pi /2 - x)} } \over {\sqrt {\cos ec(\pi /2 - x)} + \sqrt {\sec (\pi /2 - x} )}}} dx[/tex]
or [tex]I = \int\limits_0^{\pi /2} {{{\sqrt {\cos ecx} } \over {\sqrt {\sec x + } \sqrt {\cos ecx} }}} dx[/tex] ....... (2)
By (1) + (2) we get
[tex]2I = \int\limits_0^{\pi /2} {dx = {\pi \over 2}}[/tex]
There fore [tex]I = {\pi \over 4}[/tex] ....(Proved)
