If I1=∫1sin⁡θ x1+x2dx and I2=∫1cosec⁡θ 1xx2+1dx then the value of I1I12I2eI1+I2I22−11I12+I22−1, is

# If ${I}_{1}={\int }_{1}^{\mathrm{sin}\theta } \frac{x}{1+{x}^{2}}dx$ and ${I}_{2}={\int }_{1}^{\mathrm{cosec}\theta } \frac{1}{x\left({x}^{2}+1\right)}dx$ 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

1. A
2. B
3. C
4. D

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### Solution:

We have,

${I}_{2}={\int }_{1}^{\mathrm{cosec}\theta } \frac{1}{x\left({x}^{2}+1\right)}dx=-{\int }_{1}^{\mathrm{sin}\theta } \frac{t}{1+{t}^{2}}dt,$ where $t=\frac{1}{x}$

$=\left|\begin{array}{ccc}{I}_{1}& {{I}_{1}}^{2}& -{I}_{1}\\ 1& {{I}_{1}}^{2}& -1\\ 0& {{I}_{1}}^{2}& 0\end{array}\right|$                   [Applying ${R}_{3}\in {R}_{3}-{R}_{2}$]

$=-{{I}_{1}}^{2}\left(-{I}_{1}+{I}_{1}\right)=0$            [Expanding along ${R}_{3}$]

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