Question

Whichever of the following is/are correct?
(A) To evaluate \(\mathrm{I}_{1}=\int_{-2}^{2} \frac{\mathrm{dx}}{4+\mathrm{x}^{2}}\), it is possible to \(\mathrm{x}=\frac{1}{\mathrm{t}}\)
(B) To evaluate \(I_{2}=\int_{0}^{1} \sqrt{\left(x^{2}+1\right)} d x\), it is possible to put \(x=\sec t\)
(C) To evaluate \(I_{2}=\int_{0}^{1} \sqrt{\left(x^{2}+1\right)} d x\), it is not possible to put \(x=\operatorname{cosec} \theta\)
(D) To evaluate \(\mathrm{I}_{1}\), it is not possible to put \(\mathrm{x}=\frac{1}{\mathrm{t}}\)

Answer 1

Ans: (C, D)
Hint : \(I_{1}=\int_{-2}^{2} \frac{d x}{4+x^{2}}=2 \int_{0}^{2} \frac{d x}{4+x^{2}}>0\)
By
$$
x=1 / t, \quad I_{1}=\int_{-1 / 2}^{1 / 2} \frac{-d t}{4 t^{2}+1}<0
$$
So not possible
$$
I_{2}=\int_{0}^{1} \sqrt{x^{2}+1} d x, x=\operatorname{cosec} \theta
$$
Since \(x \in(0,1)\) and \(\operatorname{cosec} \theta \in(-\infty,-1] \cup[1, \infty)\)
So not possible