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WBJEE · Maths · Definite Integration

Whichever of the following is/are correct?

  1. A To evaluate \(I_{1}=\int_{-2}^{2} \frac{d x}{4+x^{2}}\), it is possible to put \(x=\frac{1}{t}\)
  2. B To evaluate \(I_{2}=\int_{0}^{1} \sqrt{\left(x^{2}+1\right)} d x\), it is possible to put \(x=\sec t\)
  3. C To evaluate \(I_{2}=\int_{0}^{1} \sqrt{\left(x^{2}+1\right)} \mathrm{dx}\), it is not possible to put \(x=\operatorname{cosec} \theta\)
  4. D To evaluate \(\mathrm{I}_{1}\), it is not possible to put \(\mathrm{x}=\frac{1}{\mathrm{t}}\)
Verified Solution

Answer & Solution

Correct Answer

(D) To evaluate \(\mathrm{I}_{1}\), it is not possible to put \(\mathrm{x}=\frac{1}{\mathrm{t}}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{I}_{1}=\int_{-2}^{2} \frac{\mathrm{dx}}{4+\mathrm{x}^{2}}=2 \int_{0}^{2} \frac{\mathrm{dx}}{4+\mathrm{x}^{2}}>0\) By \(\mathrm{x}=1 / \mathrm{t}, \quad \mathrm{I}_{1}=\int_{-1 / 2}^{1 / 2} \frac{-\mathrm{dt}}{4 \mathrm{t}^{2}+1} < 0\) So not possible…