Question

lf $f\left(x\right)=\left|\begin{array}{cc}\sec x& \cos x\\ {\cos }^{2}x& {\cos }^{2}x\end{array}\right|$, then ${\int }_0^{\pi /2}f\left(x\right)dx=$

• A. 1/2
• B. 1/3
• C. 0
• D. 1

Answer 1

The correct option is B 1/3

$f\left(x\right)=\sec x{\cos }^{2}x-\cos x\times {\cos }^{2}x.$
$=\cos x-{\cos }^{3}x.$
${\int }_0^{\pi /2}(cosx-co{s}^{3}x)dx={\int }_0^{\pi /2}\cos x{\sin }^{2}xdx$
Substitute $\sin x=t$
$\Rightarrow \cos xdx=dt$
The integral becomes
${\int }_0^1{t}^{2}dt$
$={\left[\frac{t^3}{3}\right]}_0^1=1/3$