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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2021 · Mathematics · Indefinite Integration

If ∫ 1- 2 x d x=A x+B x+C , where 0 < x < (π)/(4) , then which one...

If ∫1−sin⁡2xdx=Asin⁡x+Bcos⁡x+C\int \sqrt{1-\sin 2 x} d x=A \sin x+B \cos x+C, where 0<x<π40 < x < \frac{\pi}{4}, then which one of the following is correct?

Options

  1. AA+B=0A+B=0
  2. BA+B−2=0A+B-2=0
  3. CA+B+2=0A+B+2=0
  4. DA+B−1=0A+B-1=0

Correct answer

A. A+B=0A+B=0

Step-by-step solution

I=∫1−sin⁡2xdx,0<x<π4=∫sin⁡2x−2sin⁡xcos⁡x+cos⁡2xdx=∫(sin⁡x−cos⁡x)2dx\begin{aligned} & \mathrm{I}=\int \sqrt{1-\sin ^2 x} d x, 0 < x < \frac{\pi}{4} \\ & =\int \sqrt{\sin ^2 x-2 \sin x \cos x+\cos ^2 x} d x \\ & =\int \sqrt{(\sin x-\cos x)^2} d x \end{aligned} (i.e. sin⁡x−cos⁡x<0\sin x-\cos x < 0 when 0<x<π40 < x < \frac{\pi}{4} ) ∴I=−∫(sin⁡x−cos⁡x)dx=∫(cos⁡x−sin⁡x)dxI=sin⁡x−cos⁡x+c=Asin⁡x+Bcos⁡x+c\begin{aligned} & \therefore \mathrm{I}=-\int(\sin x-\cos x) d x=\int(\cos x-\sin x) d x \\ & \mathrm{I}=\sin x-\cos x+c=\mathrm{A} \sin x+\mathrm{B} \cos x+\mathrm{c} \end{aligned} Thus, A+B=1−1=0A+B=1-1=0
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