The integral I = ∫ ^ 3 θ cos θ (1 + ^ 2 θ) ^ 2 dθ simplifies to (where, c is the constant of integration)

Correct answer: B. (1)/(2)( (1 + ^ 2 θ) + 1 1 + ^ 2 θ ) + c Let 1 + ^ 2 θ = t ⇒ 2sin θ cos θ = dt ∴ I = (1)/(2)∫ t - 1 t^ 2 dt = (1)/(2)∫ ((1)/(t) - 1 t^ 2 )dt = (1)/(2)(ln t + (1)/(t)) + c = (1)/(2)( (1 + ^ 2 θ) + 1 1 + ^ 2 θ ) + c

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The integral I = ∫ ^ 3 θ cos θ (1 + ^ 2 θ) ^ 2 dθ simplifies to...

The integral

I = ∫ \dfrac{\sin^{3} \theta cos \theta}{{\left(1 + \sin^{2} \theta\right)}^{2}}

d\theta

simplifies to (where,

c is the constant of integration)

Options

A $\dfrac{1}{2}\ln \left(sin \theta\right) + \dfrac{1}{1 + \sin^{2} \theta} + c$
B $\dfrac{1}{2}\left(\ln \left(1 + \sin^{2} \theta\right) + \dfrac{1}{1 + \sin^{2} \theta}\right) + c$
C $\ln \left(sin \theta\right) + \dfrac{1}{1 + \sin^{2} \theta} + c$
D $\ln \left(\sin^{2} \theta + 1\right) + \dfrac{1}{\sin^{2} \theta + 2} + c$

Correct answer

B. $\dfrac{1}{2}\left(\ln \left(1 + \sin^{2} \theta\right) + \dfrac{1}{1 + \sin^{2} \theta}\right) + c$

Step-by-step solution

Let

1 + \sin^{2} \theta = t

⇒ 2sin \theta cos \thetad\theta = dt

∴ I = \dfrac{1}{2}∫ \dfrac{t - 1}{t^{2}}dt

= \dfrac{1}{2}∫ \left(\dfrac{1}{t} - \dfrac{1}{t^{2}}\right)dt

= \dfrac{1}{2}\left(ln t + \dfrac{1}{t}\right) + c

= \dfrac{1}{2}\left(\ln \left(1 + \sin^{2} \theta\right) + \dfrac{1}{1 + \sin^{2} \theta}\right) + c

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