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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If I = ∫ ( tan x + cot x )dx , then I equals to (where, C & is an...

If I=∫(tanx+cotx)dxI = ∫ \left(\sqrt{tan x} + \sqrt{cot x}\right)dx, then II equals to (where, CC& is an arbitrary constant)

Options

  1. A2sin⁡−1(sinx+cosx)+C\sqrt{2}\sin^{-1} \left(sin x + cos x\right) + C
  2. B2cos⁡−1(sinx−cosx)+C\sqrt{2}\cos^{-1} \left(sin x - cos x\right) + C
  3. C2sin⁡−1(sinx−cosx)+C\sqrt{2}\sin^{-1} \left(sin x - cos x\right) + C
  4. D2cos⁡−1(sinx+cosx)+C\sqrt{2}\cos^{-1} \left(sin x + cos x\right) + C

Correct answer

C. 2sin⁡−1(sinx−cosx)+C\sqrt{2}\sin^{-1} \left(sin x - cos x\right) + C

Step-by-step solution

I=∫sinx+cosxcosxsinxdxI = ∫ \dfrac{sin x + cos x}{\sqrt{cos xsin x}}dx

Put sinx−cosx=tsin x - cos x = t , so that \left(sin x + cos x\right)dx = dtand1–2sinxcosx=t2<br/><br/>and 1 – 2sin xcos x = t^{2} <br/><br/>∴ I = \sqrt{2}∫ \dfrac{dt}{\sqrt{1 - t^{2}}} = \sqrt{2}\sin^{- 1} \left(t\right) + C<br/><br/><br/><br/>= \sqrt{2}\sin^{- 1} \left(sin x - cos x\right) + C
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