JEE Advance - Mathematics (1987 - No. 25)
Evaluate :$$\,\,\int {\left[ {{{{{\left( {\cos 2x} \right)}^{1/2}}} \over {\sin x}}} \right]dx} $$
${1 over {sqrt 2 }},\log \left[ {{{\sqrt 2 + \sqrt {1 - {{\tan }^2}x} } \over {\sqrt 2 - \sqrt {1 - {{\tan }^2}x} }}} \right] - \log \left( {\cot x + \sqrt {{{\cot }^2}x - 1} } \right) + C
${1 over {sqrt 2 }}\,\log \left[ {{{\sqrt 2 - \sqrt {1 - {{\tan }^2}x} } \over {\sqrt 2 + \sqrt {1 - {{\tan }^2}x} }}} \right] - \log \left( {\cot x + \sqrt {{{\cot }^2}x - 1} } \right) + C
${1 over {sqrt 2 }}\,\log \left[ {{{\sqrt 2 + \sqrt {1 - {{\tan }^2}x} } \over {\sqrt 2 - \sqrt {1 - {{\tan }^2}x} }}} \right] + \log \left( {\cot x + \sqrt {{{\cot }^2}x - 1} } \right) + C
${1 over {sqrt 2 }}\,\log \left[ {{{\sqrt 2 + \sqrt {1 + {{\tan }^2}x} } \over {\sqrt 2 - \sqrt {1 + {{\tan }^2}x} }}} \right] - \log \left( {\cot x + \sqrt {{{\cot }^2}x - 1} } \right) + C
${1 over {sqrt 2 }}\,\log \left[ {{{\sqrt 2 + \sqrt {1 - {{\tan }^2}x} } \over {\sqrt 2 - \sqrt {1 - {{\tan }^2}x} }}} \right] - \log \left( {\tan x + \sqrt {{{\tan }^2}x - 1} } \right) + C
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