Question

$f\left(x\right)=\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+\cot x\text{cosec}x\\ {\cos }^{2}x& {\cos }^{2}x& {\text{cosec}}^{2}x\\ 1& {\cos }^{2}x& {\cos }^{2}x\end{array}\right|$. If $\underset{0}{\overset{\pi /2}{\int }}f\left(x\right)dx=-\left(\frac{k\pi +m}{60}\right)$, then $k+m$ is equal to

Answer 1

$f\left(x\right)=\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+\cot x\text{cosec}x\\ {\cos }^{2}x& {\cos }^{2}x& {\text{cosec}}^{2}x\\ 1& {\cos }^{2}x& {\cos }^{2}x\end{array}\right|$
Using row operation $R_1\to R_1-\sec x\cdot R_3$
$f\left(x\right)=\left|\begin{array}{ccc}0& 0& {\sec }^{2}x+\cot x\text{cosec}x-\cos x\\ {\cos }^{2}x& {\cos }^{2}x& {\text{cosec}}^{2}x\\ 1& {\cos }^{2}x& {\cos }^{2}x\end{array}\right|\Rightarrow f\left(x\right)=({\sec }^{2}x+\cot x\text{cosec}x-\cos x)({\cos }^{4}x-{\cos }^{2}x)\Rightarrow f\left(x\right)=({\sec }^{2}x+\cot x\text{cosec}x-\cos x){\cos }^{2}x({\cos }^{2}x-1)\Rightarrow f\left(x\right)=-{\sin }^{2}x(1+\frac{{\cos }^{3}x}{{\sin }^{2}x}-{\cos }^{3}x)\Rightarrow f\left(x\right)=-{\sin }^{2}x(1+{\cos }^{3}x(\frac{1}{{\sin }^{2}x}-1))\Rightarrow f\left(x\right)=-{\sin }^{2}x(1+\frac{{\cos }^{5}x}{{\sin }^{2}x})\Rightarrow f\left(x\right)=-{\cos }^{5}x-{\sin }^{2}x$

Let
$I=\underset{0}{\overset{\pi /2}{\int }}-{\cos }^{5}x-{\sin }^{2}xdx$
$\Rightarrow I=-(I_1+I_2)$
$I_1=\underset{0}{\overset{\pi /2}{\int }}{\sin }^{2}xdx\Rightarrow I_1=\underset{0}{\overset{\pi /2}{\int }}{\cos }^{2}xdx\Rightarrow 2I_1=\underset{0}{\overset{\pi /2}{\int }}1dx\Rightarrow I_1=\frac{\pi }{4}$
$I_2=\underset{0}{\overset{\pi /2}{\int }}{\cos }^{5}xdx\Rightarrow I_2=\underset{0}{\overset{\pi /2}{\int }}\cos x(1-{\sin }^{2}x{)}^{2}dx$
Put $\sin x=t\Rightarrow \cos xdx=dt$
$I_2=\underset{0}{\overset{1}{\int }}(1-t^2{)}^{2}dt\Rightarrow I_2=\underset{0}{\overset{1}{\int }}(1+t^4-2t^2)dt\Rightarrow I_2=1+\frac{1}{5}-\frac{2}{3}=\frac{8}{15}$
Therefore,
$I=-(I_1+I_2)=-\frac{\pi }{4}-\frac{8}{15}=-\left(\frac{15\pi +32}{60}\right)$
Hence, $k+m=47$