The integral I = ∫ sin (2θ) [ 1 + ^ 2 θ 2 ^ 2 θ ]dθ simplifies to (where, c is the integration constant)

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

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The integral I = ∫ sin (2θ) [ 1 + ^ 2 θ 2 ^ 2 θ ]dθ simplifies to...

The integral

I = ∫ sin

\left(2\theta\right)

\left[\dfrac{1 + \cos^{2} \theta}{2\sin^{2} \theta}\right]d\theta

simplifies to (where,

c

is the integration constant)

Options

A $\ln \left|\sin \theta\right| + cos \theta + c$
B $2\ln \left|\sin \theta\right| - \dfrac{\sin^{2} \theta}{2} + c$
C $\ln \left|\sin \theta\right| - \sin^{2} \theta + c$
D $\ln \left|\cos \theta\right| + \cos^{2} \theta + c$

Correct answer

B. $2\ln \left|\sin \theta\right| - \dfrac{\sin^{2} \theta}{2} + c$

Step-by-step solution

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

Let

sin \theta = t

⇒ cos \thetad\theta = dt

⇒ I = ∫ \dfrac{2 - t^{2}}{t}dt

= 2\ln \left|t\right| - \dfrac{t^{2}}{2} + c

= 2\ln \left|\sin \theta\right| - \dfrac{\sin^{2} \theta}{2} + c

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