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The integral $$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}} $$ is equal to

JEE · Math · previous-year question

  1. A.2correct
  2. B.4
  3. C.$$-$$ 1
  4. D.$$-$$ 2

Answer

A. 2

Explanation

$$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}} $$ = $$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {2{{\cos }^2}{x \over 2}}}} $$ = $${1 \over 2}\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{\sec }^2}} {x \over 2}\,dx$$ $${1 \over 2}\left[ {{{\tan {x \over 2}} \over {{1 \over 2}}}} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$$ = $$\left[ {\tan {x \over 2}} \right]_{{\pi \over 4}}^{{3 \over 4}}$$ = tan $${{3\pi } \over 8}$$ $$-$$ tan$${\pi \over 8}$$ = $$\left( {\sqrt 2 + 1} \right) - \left( {\sqrt 2 - 1} \right)$$ = 2

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