Which of the following is/are correct?

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(C, D) For $I_{1}=\int_{-2}^{2} \frac{dx}{4+x^{2}}$,the integrand is positive,so $I_{1} > 0$. If we substitute $x=\frac{1}{t}$,then $dx = -\frac{1}{t^

Vedclass · Accepted Answer

To evaluate $I_{1}$,it is not possible to put $x=\frac{1}{t}$

Vedclass · Answer

(C, D) For $I_{1}=\int_{-2}^{2} \frac{dx}{4+x^{2}}$,the integrand is positive,so $I_{1} > 0$. If we substitute $x=\frac{1}{t}$,then $dx = -\frac{1}{t^{2}} dt$. The limits change from $x=-2$ to $t=-1/2$ and $x=2$ to $t=1/2$. $I_{1} = \int_{-1/2}^{1/2} \frac{-dt/t^{2}}{4+1/t^{2}} = \int_{-1/2}^{1/2} \frac{-dt}{4t^{2}+1}$. Since the integrand is positive,the integral must be positive,but this substitution leads to a negative value,which is incorrect due to the discontinuity of $1/t$ at $t=0$. Thus,it is not possible. For $I_{2}=\int_{0}^{1} \sqrt{x^{2}+1} dx$,if we put $x=\operatorname{cosec} 	heta$,then $\operatorname{cosec} 	heta \in (-\infty, -1] \cup [1, \infty)$. Since $x \in (0, 1)$,the range of $x$ does not match the range of $\operatorname{cosec} 	heta$. Thus,it is not possible. Therefore,options $C$ and $D$ are correct.

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