MCQ
$\int_{-\pi / 4}^{\pi / 4} \sec ^2 x d x$ is equal to
  • A
    $-1$
  • B
    $0$
  • C
    $1$
  • $2$

Answer

Correct option: D.
$2$
$\text {Let } I=\int_{-\pi / 4}^{\pi / 4} \sec ^2 x d x$
$=[\tan x]_{-\pi / 4}^{\pi / 4}$
$=\tan \frac{\pi}{4}-\tan \left(-\frac{\pi}{4}\right)$
$=1+1$
$=2$

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