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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS1 Mark
Question
$\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot^3\text{x}}\text{ dx}$ is equal to:
  1. $0$
  2. $1$
  3. $\frac{\pi}{2}$
  4. $\frac{\pi}{4}$
✓

Answer

  1. $\frac{\pi}{4}$
Solution:
We have,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot^3\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot^3\big(\frac{\pi}{2}-{\text{x}}\big)}\text{ dx}$
$\therefore\ \text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\tan^3\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big[\frac{1}{1+\cot^3\text{x}}+\frac{1}{\tan^3\text{x}}\Big]\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\bigg[\frac{2+\tan^3\text{x}+1+\cot^3\text{x}}{(1+\cot^3\text{x})(1+\tan^3\text{x})}\bigg]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\bigg[\frac{2\tan^3\text{x}+\cot^3\text{x}}{1+\tan^3\text{x}+\cot^3\text{x}+\cot^3\text{x}\tan^3\text{x}}\bigg]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\Big[\frac{2\tan^3\text{x}+\cot^3\text{x}}{1+\tan^3\text{x}+\cot^3\text{x}+1}\Big]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\Big[\frac{2+\tan^3\text{x}+\cot^3\text{x}}{2+\tan^3\text{x}+\cot^3\text{x}}\Big]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\Big[1\Big]^{\frac{\pi}{2}}_0$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0$
$=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$

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