If {I_1} = intlimits_1^{sin theta } {frac{x}{ | vedclass

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(D) Given ${I_1} = \int_1^{\sin 	heta } {\frac{x}{{1 + x^2}}} dx$ and ${I_2} = \int_1^{\csc 	heta } {\frac{1}{{x\left( {{x^2} + 1} ight)}}} dx$.Fo

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(D) Given ${I_1} = \int_1^{\sin 	heta } {\frac{x}{{1 + x^2}}} dx$ and ${I_2} = \int_1^{\csc 	heta } {\frac{1}{{x\left( {{x^2} + 1} ight)}}} dx$.For ${I_2}$,substitute $x = \frac{1}{t}$,so $dx = -\frac{1}{t^2} dt$.When $x=1, t=1$. When $x=\csc 	heta, t=\sin 	heta$.${I_2} = \int_1^{\sin 	heta } {\frac{{ - \frac{1}{t^2} dt}}{{\frac{1}{t}\left( {\frac{1}{t^2} + 1} ight)}}} = \int_1^{\sin 	heta } {\frac{{ - \frac{1}{t^2} dt}}{{\frac{1}{t}\left( {\frac{{1 + t^2}}{{t^2}}} ight)}}} = \int_1^{\sin 	heta } {\frac{{ - t dt}}{{1 + t^2}}} = - \int_1^{\sin 	heta } {\frac{t}{{1 + t^2}}} dt = -{I_1}$.Thus,${I_1} + {I_2} = 0$.Now,consider the determinant $\Delta = \left| {\begin{array}{*{20}{c}} {{I_1}}&{I_1^2}&{{I_2}} \ {{e^{{I_1} + {I_2}}}}&{I_2^2}&{ - 1} \ 1&{I_1^2 + I_2^2}&{ - 1} \end{array}} ight|$.Since ${I_1} + {I_2} = 0$,$e^{{I_1} + {I_2}} = e^0 = 1$.$\Delta = \left| {\begin{array}{*{20}{c}} {{I_1}}&{I_1^2}&{{I_2}} \ 1&{I_2^2}&{ - 1} \ 1&{I_1^2 + I_2^2}&{ - 1} \end{array}} ight|$.Applying the operation $C_1 ightarrow C_1 + C_3$,we get:$\Delta = \left| {\begin{array}{*{20}{c}} {{I_1} + {I_2}}&{I_1^2}&{{I_2}} \ {1 - 1}&{I_2^2}&{ - 1} \ {1 - 1}&{I_1^2 + I_2^2}&{ - 1} \end{array}} ight| = \left| {\begin{array}{*{20}{c}} 0&{I_1^2}&{{I_2}} \ 0&{I_2^2}&{ - 1} \ 0&{I_1^2 + I_2^2}&{ - 1} \end{array}} ight|$.Since the first column is all zeros,the value of the determinant is $0$.

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