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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The integral ∫ x cos ^ - 1 ( 1 - x ^ 2 1 + x ^ 2 ) dx; (x > 0) is...

The integral ∫xcos−1∫ \text{x} {\text{cos}}^{- 1}\left(\dfrac{1 - {\text{x}}^{2}}{1 + {\text{x}}^{2}}\right) \text{dx}; (x>0)\left(\text{x} > 0\right) is equal to

Options

  1. A−x+(1+x2)cot−1x+c- \text{x} + \left(1 + {\text{x}}^{2}\right) {\text{cot}}^{- 1} \text{x} + \text{c}
  2. Bx−(1+x2)cot−1x+c\text{x} - \left(1 + {\text{x}}^{2}\right) {\text{cot}}^{- 1} \text{x} + \text{c}
  3. Cx−(1+x2)tan−1x+c\text{x} - \left(1 + {\text{x}}^{2}\right) {\text{tan}}^{- 1} \text{x} + \text{c}
  4. D−x+(1+x2)tan−1x+c- \text{x} + \left(1 + {\text{x}}^{2}\right) {\text{tan}}^{- 1} \text{x} + \text{c}

Correct answer

D. −x+(1+x2)tan−1x+c- \text{x} + \left(1 + {\text{x}}^{2}\right) {\text{tan}}^{- 1} \text{x} + \text{c}

Step-by-step solution

I=∫xcos⁡−1(1−x21+x2)dx;(x>0)\text{I} = ∫ \text{x} \cos^{-1} \left(\dfrac{1 - {\text{x}}^{2}}{1 + {\text{x}}^{2}}\right) dx ; (x > 0)

Let,x=tan\theta⇒dx=sec2\thetad\theta\text{Let,} x = \text{tan} \text{\theta} ⇒ dx = {\text{sec}}^{2} \text{\theta} d \text{\theta}

cos−1(1−x21+x2)=cos−1cos⁡2\theta{\text{cos}}^{-1} \left(\dfrac{1 - {\text{x}}^{2}}{1 + {\text{x}}^{2}}\right) = {\text{cos}}^{-1} \cos 2 \text{\theta}

=2\theta= 2 \text{\theta}

=2tan−1x= 2 {\text{tan}}^{-1} \text{x}

I=∫(tan\theta)2\theta⋅sec2\thetad\theta\text{I} = ∫ \left(\text{tan} \text{\theta}\right) 2 \text{\theta} \cdot {\text{sec}}^{2} \text{\theta} \text{d} \text{\theta}

=2[\theta∫tan\thetasec⁡2\thetad\theta−∫(∫tan\thetasec2\thetad\theta)(dd\theta\theta)d\theta]= 2 \left[\text{\theta} ∫ \text{tan} \text{\theta} \sec^{2} \text{\theta} \text{d} \text{\theta} - ∫ \left(∫ \text{tan} \text{\theta} {\text{sec}}^{2} \text{\theta} \text{d} \text{\theta}\right) \left(\dfrac{\text{d}}{\text{d} \text{\theta}} \text{\theta}\right) \text{d} \text{\theta}\right]

=2[\thetatan2\theta2−∫tan2\theta21d\theta]= 2 \left[\text{\theta} \dfrac{{\text{tan}}^{2} \text{\theta}}{2} - ∫ \dfrac{{\text{tan}}^{2} \text{\theta}}{2} 1 \text{d} \text{\theta}\right]

=22[\thetatan2\theta−∫(sec2\theta−1)d\theta]= \dfrac{2}{2} \left[\text{\theta} {\text{tan}}^{2} \text{\theta} - ∫ \left({\text{sec}}^{2} \text{\theta} - 1\right) \text{d} \text{\theta}\right]

=\thetatan2\theta−tan\theta+\theta+c= \text{\theta} {\text{tan}}^{2} \text{\theta} - \text{tan} \text{\theta} + \text{\theta} + \text{c}

=x2tan−1x−x+tan−1x+c= {\text{x}}^{2} {\text{tan}}^{-1} \text{x} - \text{x} + {\text{tan}}^{-1} \text{x} + \text{c}

=−x+(1+x2)tan−1x+c= - \text{x} + \left(1 + {\text{x}}^{2}\right) {\text{tan}}^{-1} \text{x} + \text{c}
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