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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
Choose the correct answer in Exercise : The value of $\int^{1}_{0}\tan^{-1}\bigg(\frac{2\text{x}-1}{1+\text{x}-\text{x}^{2}}\bigg)\text{dx}$ is
  • A
    $1$
  • ✓
    $0$
  • C
    $-1$
  • D
    $\frac{\pi}{4}$
✓

Answer

Correct option: B.
$0$
$\text{Let I}=\int^{1}\limits_{0}\tan^{-1}\bigg(\frac{2\text{x}-1}{1+\text{x}-\text{x}^{2}}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int^{1}\limits_{0}\tan^{-1}\bigg(\frac{\text{x}-(1-\text{x})}{1+\text{x}(1-\text{x})}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int^{1}\limits_{0}\Big[\tan^{-1}\text{x}-\tan^{-1}(1-\text{x)}\Big]\text{dx}$
$\Rightarrow\text{I}=\int^{\text{1}}\limits_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}(1-1+\text{x)}\Big]\text{dx}$
$\Rightarrow\text{I}=\int^{1}\limits_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}\text{(x)}\Big]\text{dx}$
$\Rightarrow\text{I}=\int^{1}_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}\text{(x)}\Big]\text{dx}$
Adding $(1)$ and $(2),$ we obtain
$\Rightarrow2\text{I}=\int^{1}\limits_{0}\Big(\tan^{-1}\text{x)}+\tan^{-1}(1-\text{x)}-\tan^{-1}\text{x}\Big)\text{dx}$
$\Rightarrow2\text{I}=0$
$\Rightarrow\text{I}=0$

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