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

Solution:

We have,

$I_{2} = \int_{1}^{cosec θ} \frac{1}{x (x^{2} + 1)} d x = - \int_{1}^{\sin θ} \frac{t}{1 + t^{2}} d t,$ where $t = \frac{1}{x}$

$\begin{array}{l} \Rightarrow I_{2} = - I_{1} \\ ∴ |\begin{array}{ccc} I_{1} & {I_{1}}^{2} & I_{2} \\ e^{I_{1} + I_{2}} & {I_{2}}^{2} & - 1 \\ 1 & {I_{1}}^{2} + {I_{2}}^{2} & - 1 \end{array}| = |\begin{array}{ccc} I_{1} & {I_{1}}^{2} & - I_{1} \\ e^{0} & {I_{1}}^{2} & - 1 \\ 1 & 2 {I_{1}}^{2} & - 1 \end{array}| \end{array}$

$= |\begin{array}{ccc} I_{1} & {I_{1}}^{2} & - I_{1} \\ 1 & {I_{1}}^{2} & - 1 \\ 0 & {I_{1}}^{2} & 0 \end{array}|$ [Applying $R_{3} \in R_{3} - R_{2}$ ]

$= - {I_{1}}^{2} (- I_{1} + I_{1}) = 0$ [Expanding along $R_{3}$ ]

If $I_{1} = \int_{1}^{\sin θ} \frac{x}{1 + x^{2}} d x$ and $I_{2} = \int_{1}^{cosec θ} \frac{1}{x (x^{2} + 1)} d x$ then the value of $|\begin{array}{ccc} I_{1} & {I_{1}}^{2} & I_{2} \\ e^{I_{1} + I_{2}} & {I_{2}}^{2} & - 1 \\ 1 & {I_{1}}^{2} + {I_{2}}^{2} & - 1 \end{array}|$ , is

Solution:

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