Verified Solution • 2025–26 Session moderate

QJEEM04Ma74f6: JEE Main 2004 Mathematics - Class 12 JEEM04Ma74f6 Solution

\int {{{dx} \over {\cos x - \sin x}}}

is equal to

Multiple Choice Options

A ${1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} + {{3\pi } \over 8}} \right)} \right| + C$
B ${1 \over {\sqrt 2 }}\log \left| {\cot \left( {{x \over 2}} \right)} \right| + C$
C ${1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} - {{3\pi } \over 8}} \right)} \right| + C$
D $\,{1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} - {\pi \over 8}} \right)} \right| + C$

Step-by-Step Verified Solution

\int {{{dx} \over {\cos x - \sin x}}}

= \int {{{dx} \over {\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)}}}

= {1 \over {\sqrt 2 }}\int {\sec \left( {x + {\pi \over 4}} \right)dx}

= {1 \over {\sqrt 2 }}\log \left| {\tan \left( {{\pi \over 4} + {x \over 2} + {\pi \over 8}} \right)} \right| + C

\left[ \, \right.

As

\int {\sec x\,dx} = \log \left| {\tan \left( {{\pi \over 4} + {x \over 2}} \right)} \right|

\left. \, \right]

= {1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} + {{3\pi } \over 8}} \right)} \right| + C

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