If ∫ e^ sin θ (sin θ + ^ 2 θ)dθ is equal to f(θ) + C (where, C is the constant of integration) and f (0) = 0 , then the value of f ((π)/(4)) is

Correct answer: B. e^ 1 2 ∫ e^ sin θ sin θ +∫ e^ sin θ ^ 2 θ ∫ e^ sin θ sin θ +(tan ^ sin θ ) - ∫ tan ^ sin θ cos θ = ∫ e^ sin θ sin θ + (tan θ)e^ sin θ - ∫ e^ sin θ sin θ = e^ sin θ (tan θ) + C = e^ sin θ tan θ + C f(θ) = e^ sin θ tan θ f((π)/(4)) = e^ 1 2

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If ∫ e^ sin θ (sin θ + ^ 2 θ)dθ is equal to f(θ) + C (where, C is...

∫ e^{sin \theta}\left(sin \theta + \sec^{2} \theta\right)d\theta

is equal to

f\left(\theta\right) + C

(where,

C is the constant of integration)

and f \left(0\right) = 0

, then the value of f \left(\dfrac{\pi}{4}\right)

Options

A $e^{\sqrt{2}}$
B $e^{\dfrac{1}{\sqrt{2}}}$
C $e^{2}$
D $e^{\dfrac{1}{2}}$

Correct answer

B. $e^{\dfrac{1}{\sqrt{2}}}$

Step-by-step solution

∫ e^{sin \theta}sin \thetad\theta +∫ e^{sin \theta}\sec^{2} \thetad\theta

∫ e^{sin \theta}sin \thetad\theta +\left(tan \thetae^{sin \theta}\right) - ∫ tan \thetae^{sin \theta}cos \thetad\theta

= ∫ e^{sin \theta}sin \thetad\theta + \left(tan \theta\right)e^{sin \theta} - ∫ e^{sin \theta}sin \thetad\theta = e^{sin \theta}\left(tan \theta\right) + C

= e^{sin \theta}tan \theta + C

f\left(\theta\right) = e^{sin \theta}tan \theta

f\left(\dfrac{\pi}{4}\right) = e^{\dfrac{1}{\sqrt{2}}}

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