JEE MAIN - Mathematics (2004 - No. 17)
$$\int {{{dx} \over {\cos x - \sin x}}} $$ is equal to
$${1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} + {{3\pi } \over 8}} \right)} \right| + C$$
$${1 \over {\sqrt 2 }}\log \left| {\cot \left( {{x \over 2}} \right)} \right| + C$$
$${1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} - {{3\pi } \over 8}} \right)} \right| + C$$
$$\,{1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} - {\pi \over 8}} \right)} \right| + C$$
Explanation
$$\int {{{dx} \over {\cos x - \sin x}}} $$
$$ = \int {{{dx} \over {\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)}}} $$
$$ = {1 \over {\sqrt 2 }}\int {\sec \left( {x + {\pi \over 4}} \right)dx} $$
$$ = {1 \over {\sqrt 2 }}\log \left| {\tan \left( {{\pi \over 4} + {x \over 2} + {\pi \over 8}} \right)} \right| + C$$
$$\left[ \, \right.$$ As $$\int {\sec x\,dx} = \log \left| {\tan \left( {{\pi \over 4} + {x \over 2}} \right)} \right|$$ $$\left. \, \right]$$
$$ = {1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} + {{3\pi } \over 8}} \right)} \right| + C$$
$$ = \int {{{dx} \over {\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)}}} $$
$$ = {1 \over {\sqrt 2 }}\int {\sec \left( {x + {\pi \over 4}} \right)dx} $$
$$ = {1 \over {\sqrt 2 }}\log \left| {\tan \left( {{\pi \over 4} + {x \over 2} + {\pi \over 8}} \right)} \right| + C$$
$$\left[ \, \right.$$ 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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