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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 ∫ ( tan ^ - 1 (sin x + 1))cos x (3 + 2sin x - cos^ 2...

The value of ∫( tan −1(sinx+1))cosx(3+2sinx−cos2x)dx∫ \dfrac{\left(\text{ tan }^{- 1}\left(sin x + 1\right)\right)cos x}{\left(3 + 2sin x - cos^{2}x\right)}dx is (where cc& is the constant of integration)

Options

  1. Atan−1(sinx)+ctan^{- 1}\left(sin x\right) + c
  2. B(tan−1(sinx))2+c{\left(tan^{- 1}\left(sin x\right)\right)}^{2} + c
  3. C(tan−1(sinx+1))22+c\dfrac{{\left(tan^{- 1}\left(sin x + 1\right)\right)}^{2}}{2} + c
  4. D(tan−1(sinx))22+c\dfrac{{\left(tan^{- 1}\left(sin x\right)\right)}^{2}}{2} + c

Correct answer

C. (tan−1(sinx+1))22+c\dfrac{{\left(tan^{- 1}\left(sin x + 1\right)\right)}^{2}}{2} + c

Step-by-step solution

Let, I=∫tan−1(sinx+1)cosx3+2sinx−(1−sin2x)dxΙ = ∫ \dfrac{tan^{- 1}\left(sin x + 1\right)cos x}{3 + 2sin x - \left(1 - sin^{2}x\right)}dx

Substituting, 1+sinx=t⇒cosxdx=dt1 + sin x = t ⇒ cos xdx = dt

I=∫tan−1(t)dt2+(t−1)2+2(t−1)Ι = ∫ \dfrac{tan^{- 1}\left(t\right)dt}{2 + {\left(t - 1\right)}^{2} + 2\left(t - 1\right)}

I=∫tan−1(t)dt1+t2Ι = ∫ \dfrac{tan^{- 1}\left(t\right)dt}{1 + t^{2}}

I=(tan−1(t))22+cΙ = \dfrac{{\left(tan^{- 1}\left(t\right)\right)}^{2}}{2} + c

I=(tan−1(sinx+1))22+cΙ = \dfrac{{\left(tan^{- 1}\left(sin x + 1\right)\right)}^{2}}{2} + c
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