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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2023 · Paper 2 · Mathematics · Definite Integration

∫_0^ π / 4 x e^ x d x is equal to

∫0π/4cos⁡xesin⁡xdx\int_0^{\pi / 4} \cos x e^{\sin x} d x is equal to

Options

  1. Ae+1e+1
  2. Be−1e-1
  3. Cee
  4. D−e-e

Correct answer

B. e−1e-1

Step-by-step solution

Suppose, I=∫0π/2cos⁡xesin⁡xdx<br/>LetI=\int_0^{\pi / 2} \cos x e^{\sin x} d x<br />Let\sin x=t \Rightarrow \cos x d x=d t<br/><br />x \rightarrow 0 \Rightarrow t \rightarrow 0<br/>and<br />andx \rightarrow \frac{\pi}{2} \Rightarrow t \rightarrow 1<br/>So<br />SoI=\int_0^1 e^t d t=\left[e^t\right]_0^1=e-1
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