MCQ
$\int_{ - \pi /2}^{\pi /2} {{{\sin }^2}x{{\cos }^2}x(\sin x + \cos x)\,dx = } $
- A$\frac{2}{{15}}$
- ✓$\frac{4}{{15}}$
- C$\frac{6}{{15}}$
- D$\frac{8}{{15}}$
$= \int_{ - \pi /2}^{\pi /2} {{{\sin }^3}x{{\cos }^2}xdx + \int_{ - \pi /2}^{\pi /2} {{{\sin }^2}x{{\cos }^3}x\,dx} } $
$ = 0 + 2\int_0^{\pi /2} {{{\sin }^2}x{{\cos }^3}xdx} $
$ = 0 + 2 \times \frac{2}{{15}} = \frac{4}{{15}}$ .
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$\left| {\begin{array}{*{20}{c}}
{ - 1 + \cos B}&{\cos C + \cos B}&{\cos B} \\
{\cos C + \cos A}&{ - 1 + \cos A}&{\cos A} \\
{ - 1 + \cos B}&{ - 1 + \cos A}&{ - 1}
\end{array}} \right|$ ની કિમંત મેળવો.