Question

If $f\left(x\right)=\left|\begin{array}{ccc}\tan x& \sin x\cos x& \tan x\sec x+\cos x\\ {\cos }^{2}x& {\cos }^{2}x& {cosec}^{2}x\\ 1& {\cos }^{2}x& {\cos }^{2}x\end{array}\right|$ Then $\underset{+\pi }{\overset{-\pi }{\int }}f\left(x\right)dx=2k\pi$. Therefore, the value of $k$ is

Answer 1

$R_3\to R_3-\left(\cot x\right)R_1$
$f\left(x\right)=\left|\begin{array}{ccc}\tan x& \sin x\cos x& \tan x\sec x+\cos x\\ {\cos }^{2}x& {\cos }^{2}x& {cosec}^{2}x\\ 0& 0& {\cos }^{2}x-\sec x-\frac{{\cos }^{2}x}{\sin x}\end{array}\right|$
$\Rightarrow f\left(x\right)=(\sin x\cos x-{\cos }^{3}x\sin x)({\cos }^{2}x-\sec x-\frac{{\cos }^{2}x}{\sin x})$
We know that,
$(\sin x\cos x-{\cos }^{3}x\sin x)({\cos }^{2}x-\sec x)$ is an even function so,
$\underset{+\pi }{\overset{-\pi }{\int }}f\left(x\right)dx=\underset{+\pi }{\overset{-\pi }{\int }}-{\cos }^{3}x-{\cos }^{5}xdx\Rightarrow I=\underset{+\pi }{\overset{-\pi }{\int }}-{\cos }^{3}x-{\cos }^{5}xdx\Rightarrow I=\underset{-\pi }{\overset{+\pi }{\int }}{\cos }^{3}x+{\cos }^{5}xdx\Rightarrow I=2\underset{0}{\overset{+\pi }{\int }}{\cos }^{3}x+{\cos }^{5}xdx$
Substituting $x\to \pi -x$
$\Rightarrow I=2\underset{0}{\overset{+\pi }{\int }}-{\cos }^{3}x-{\cos }^{5}xdx\Rightarrow I=0$
$\Rightarrow k=0$