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BITSAT 2024Paper 1
Exam
BITSAT
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Indefinite Integration

BITSAT 2024 · Paper 1 · Mathematics · Indefinite Integration

The value of ∫ e^ θ ( θ- θ) d θ is

The value of ∫etan⁡θ(sec⁡θ−sin⁡θ)dθ\int e^{\tan \theta}(\sec \theta-\sin \theta) d \theta is

Options

  1. Aetan⁡θsec⁡θ+ce^{\tan \theta} \sec \theta+c
  2. Betan⁡θsin⁡θ+ce^{\tan \theta} \sin \theta+c
  3. Cetan⁡θ(sec⁡θ+sin⁡θ)+ce^{\tan \theta}(\sec \theta+\sin \theta)+c
  4. Detan⁡θcos⁡θ+ce^{\tan \theta} \cos \theta+c

Correct answer

D. etan⁡θcos⁡θ+ce^{\tan \theta} \cos \theta+c

Step-by-step solution

Let I=∫etan⁡θ(sec⁡θ−sin⁡θ)dθI=\int e^{\tan \theta}(\sec \theta-\sin \theta) d \theta Put tan⁡θ=t⇒sec⁡2θdθ=dt⇒dθ=dt1+t2⇒I=∫et(1+t2−t1+t2)dt1+t2=∫et(11+t2−t(1+t2)3/2)dt\tan \theta=t \Rightarrow \sec ^2 \theta d \theta=d t \Rightarrow d \theta=\frac{d t}{1+t^2}\Rightarrow I=\int e^t\left(\sqrt{1+t^2}-\frac{t}{\sqrt{1+t^2}}\right) \frac{d t}{1+t^2}=\int e^t\left(\frac{1}{\sqrt{1+t^2}}-\frac{t}{\left(1+t^2\right)^{3 / 2}}\right) d t Integrating first part by parts we have, =11+t2et+∫t(1+t2)3/2⋅etdt−∫t(1+t2)3/2etdt+c=et1+t2+c=etan⁡θcos⁡θ+c=\frac{1}{\sqrt{1+t^2}} e^t+\int \frac{t}{\left(1+t^2\right)^{3 / 2}} \cdot e^t d t-\int \frac{t}{\left(1+t^2\right)^{3 / 2}} e^t d t+c=\frac{e^t}{\sqrt{1+t^2}}+c=e^{\tan \theta} \cos \theta+c
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