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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The value of the integral ∫ sin^ 2 xsec^ 2 x + 2 1 - x^ 2 x ^ - 1 x...

The value of the integral ∫sin2xsec2x+21−x2tan⁡xsin⁡−1x1−x2(1+ tan 2x)∫ \dfrac{sin^{2}xsec^{2}x + 2\sqrt{1 - x^{2}}\tan x \sin^{- 1}x}{\sqrt{1 - x^{2}}\left(1 +\text{ tan }^{2}x\right)}dxis(where,is (where,C is the constant of integration)

Options

  1. A(sin−1x)(cos2x)+C\left(sin^{- 1}x\right)\left(cos^{2}x\right) + C
  2. B(sin−1x)(sin2x)+C\left(sin^{- 1}x\right)\left(sin^{2}x\right) + C
  3. C(cos−1x)(sin2x)+C\left(cos^{- 1}x\right)\left(sin^{2}x\right) + C
  4. D−sin−1x(sin2x)+C- sin^{- 1}x\left(sin^{2}x\right) + C

Correct answer

B. (sin−1x)(sin2x)+C\left(sin^{- 1}x\right)\left(sin^{2}x\right) + C

Step-by-step solution

∫(sin2x1−x2+2tan⁡x(1+tan2x)sin−1x)dx∫ \left(\dfrac{sin^{2}x}{\sqrt{1 - x^{2}}} + \dfrac{2\tan x}{\left(1 + tan^{2}x\right)}sin^{- 1}x\right)dx
=∫(sin2x1−x2+(sin⁡2x)(sin⁡−1x))dx= ∫ \left(\dfrac{sin^{2}x}{1 - x^{2}} + \left(\sin 2x\right)\left(\sin^{-1}x\right)\right)dx
=(sin⁡2x)(sin−1x)+C= \left(\sin^{2} x\right)\left(sin^{- 1}x\right) + C
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