_0^ /4 ^2 x ,dx = - Vidyadip

MCQ

$\int_0^{\pi /4} {{{\tan }^2}x\,dx = } $

✓
$1 - \frac{\pi }{4}$
B
$1 + \frac{\pi }{4}$
C
$\frac{\pi }{4} - 1$
D
$\frac{\pi }{4}$

✓

Answer

Correct option: A.

$1 - \frac{\pi }{4}$

a
(a) $\int_0^{\pi /4} {{{\tan }^2}xdx = \int_0^{\pi /4} {({{\sec }^2}x - 1)dx} } $

$ = \int_0^{\pi /4} {{{\sec }^2}xdx - \int_0^{\pi /4} {\,\,1dx} } $

$= [\tan x]_0^{\pi /4} - [x]_0^{\pi /4} = 1 - \frac{\pi }{4}$.

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