JEE MAIN - Mathematics (2017 (Offline) - No. 16)
The integral $$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}} $$ is equal to
2
4
$$-$$ 1
$$-$$ 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
= $$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {2{{\cos }^2}{x \over 2}}}} $$
= $${1 \over 2}\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{\sec }^2}} {x \over 2}\,dx$$
$${1 \over 2}\left[ {{{\tan {x \over 2}} \over {{1 \over 2}}}} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$$
= $$\left[ {\tan {x \over 2}} \right]_{{\pi \over 4}}^{{3 \over 4}}$$
= tan $${{3\pi } \over 8}$$ $$-$$ tan$${\pi \over 8}$$
= $$\left( {\sqrt 2 + 1} \right) - \left( {\sqrt 2 - 1} \right)$$
= 2
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