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

MCQ

$\int_0^{\pi /4} {\frac{{dx}}{{{{\cos }^4}x - {{\cos }^2}x{{\sin }^2}x + {{\sin }^4}x}} = } $

✓
$\frac{\pi }{2}$
B
$\frac{\pi }{4}$
C
$\frac{\pi }{3}$
D
None of these

✓

Answer

Correct option: A.

$\frac{\pi }{2}$

a
(a) Divide ${N^r}$ and ${D^r}$ by ${\cos ^4}x$

$\therefore$ $I = \int_0^{\pi /4} {\frac{{{{\sec }^2}x\,{{\sec }^2}x\,dx}}{{1 - {{\tan }^2}x + {{\tan }^4}x}}} $

Put $\tan x = t$ and ${\sec ^2}xdx = dt$

and adjust the limits, we get $I = \int_0^1 {\frac{{(1 + {t^2})}}{{{t^4} - {t^2} + 1}}} \,dt$

$ = \left[ {{{\tan }^{ - 1}}\frac{{{t^2} - 1}}{t}} \right]_0^1 $

$= {\tan ^{ - 1}}(0) - {\tan ^{ - 1}}( - \infty ) = \frac{\pi }{2}$.

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