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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
$\int\limits^\frac{\pi}{2}_0\frac{1}{1+\tan\text{x}}\text{dx}$ is equal to:
  • ✓
    $\frac{\pi}{4}$
  • B
    $\frac{\pi}{3}$
  • C
    $\frac{\pi}{2}$
  • D
    $\pi$
✓

Answer

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

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