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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Mathematics
Chapter
Indefinite Integration

BITSAT 2023 · Paper 1 · Mathematics · Indefinite Integration

If ∫ ^ x (1+ x) dx 1+ x = ^ x f(x)+C , then f(x) is equal to

If ∫\mathrmex(1+sin⁡x)dx1+cos⁡x=\mathrmexf(x)+C\int \frac{\mathrme^{\mathrm{x}}(1+\sin \mathrm{x}) \mathrm{dx}}{1+\cos \mathrm{x}}=\mathrme^{\mathrm{x}} \mathrm{f}(\mathrm{x})+\mathrm{C}, then f(x)\mathrm{f}(\mathrm{x}) is equal to

Options

  1. Asin⁡x2\sin \frac{x}{2}
  2. Bcos⁡x2\cos \frac{x}{2}
  3. Ctan⁡x2\tan \frac{x}{2}
  4. Dlog⁡x2\log \frac{x}{2}

Correct answer

C. tan⁡x2\tan \frac{x}{2}

Step-by-step solution

∫ex(1+sin⁡x)(1+cos⁡x)dx\int e^x \frac{(1+\sin x)}{(1+\cos x)} d x
=∫ex[12sec⁡2x2+tan⁡x2]dx=\int e^x\left[\frac{1}{2} \sec ^2 \frac{x}{2}+\tan \frac{x}{2}\right] d x
=12∫exsec⁡2x2dx+∫extan⁡x2dx=\frac{1}{2} \int e^x \sec ^2 \frac{x}{2} d x+\int e^x \tan \frac{x}{2} d x
=\mathrmextan⁡x2+C=\mathrme^{\mathrm{x}} \tan \frac{\mathrm{x}}{2}+\mathrm{C}
But I=\mathrmexf(x)+C\mathrm{I}=\mathrme^{\mathrm{x}} \mathrm{f}(\mathrm{x})+\mathrm{C} (given)
∴f(x)=tan⁡x2\therefore \mathrm{f}(\mathrm{x})=\tan \frac{\mathrm{x}}{2}
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