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$\frac{\pi }{3}$
- Dએકપણ નહીં.
$\therefore$ $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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