MCQ
$\int_0^{\pi /6} {\frac{{\sin x}}{{{{\cos }^3}x}}\,dx = } $
- A$\frac{2}{3}$
- ✓$\frac{1}{6}$
- C$2$
- D$\frac{1}{3}$
Put $t = \tan x $
$\Rightarrow dt = {\sec ^2}x\,dx,$ then we have
$I = \int_0^{\frac{1}{{\sqrt 3 }}} {t\,dt = } \left[ {\frac{{{t^2}}}{2}} \right]_0^{\frac{1}{{\sqrt 3 }}} = \frac{1}{6}$.
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$g(\alpha)=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin ^{\alpha} x}{\cos ^{\alpha} x+\sin ^{\alpha} x} d x$ આપેલ છે .