If $I_1=\int\limits_1^{\sin \theta} \frac{x}{1+x^2} d x$ and $I_2=\int\limits_1^{cosec \theta} \frac{1}{x\left(x^2+1\right)} d x$ then the value of $\left|\begin{array}{ccc}I_1 & I_1{ }^2 & I_2 \\ e^{I_1+I_2} & I_2{ }^2 & -1 \\ 1 & I_1{ }^2+I_2{ }^2 & -1\end{array}\right|$, is

0

We have,

$I_2=\int\limits_1^{cosec \theta} \frac{x}{x\left(x^2+1\right)} d x=-\int\limits_1^{\sin \theta} \frac{t}{1+t^2} d t$, where $t=\frac{1}{x}$

$\Rightarrow I_2=-I_1$

∴  $\left|\begin{array}{ccc} I_1 & I_1{ }^2 & I_2 \\ e^{I_1+I_2} & I_2{ }^2 & -1 \\ 1 & I_1{ }^2+I_2{ }^2 & -1 \end{array}\right|=\left|\begin{array}{rrr} I_1 & I_1{ }^2 & -I_1 \\ e^0 & I_1{ }^2 & -1 \\ 1 & 2 I_1{ }^2 & -1 \end{array}\right|$

$=\left|\begin{array}{ccc}I_1 & I_1{ }^2 & -I_1 \\ 1 & I_1^2 & -1 \\ 0 & I_1^2 & 0\end{array}\right|$        [Applying R₃ ∈ R₃ - R₂]

$=-I_1{ }^2\left(-I_1+I_1\right)=0$               [Expanding alone R₃]