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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Mathematics
Chapter
Indefinite Integration

BITSAT 2021 · Mathematics · Indefinite Integration

The value of ∫ ( x) d x is :

The value of ∫cos⁡(log⁡x)dx\int \cos (\log x) \mathrm{d} \mathrm{x} is :

Options

  1. A12[sin⁡(log⁡x)+cos⁡(log⁡x)]+C\frac{1}{2}[\sin (\log x)+\cos (\log x)]+C
  2. Bx2[sin⁡(log⁡x)+cos⁡(log⁡x)]+C\frac{x}{2}[\sin (\log x)+\cos (\log x)]+C
  3. Cx2[sin⁡(log⁡x)−cos⁡(log⁡x)]+C\frac{x}{2}[\sin (\log x)-\cos (\log x)]+C
  4. D12[sin⁡(log⁡x)−cos⁡(log⁡x)]+C\frac{1}{2}[\sin (\log x)-\cos (\log x)]+C

Correct answer

B. x2[sin⁡(log⁡x)+cos⁡(log⁡x)]+C\frac{x}{2}[\sin (\log x)+\cos (\log x)]+C

Step-by-step solution

Let (I\mathrm{I}=\int \cos (\log \mathrm{x}) \cdot 1 \mathrm{~d} \mathrm{x} \ldots . . \text{ (i) }) Use integral by parts, (I=cos⁡(log⁡x)⋅x−∫[−sin⁡(log⁡x)]⋅1x⋅xdx=x⋅cos⁡(log⁡x)+∫sin⁡(log⁡x)⋅1dx=x⋅cos⁡(log⁡x)+[sin⁡(log⁡x)⋅x−∫cos⁡(log⁡x)⋅1x⋅xdx]+C=x⋅(log⁡x)+[xsin⁡(log⁡x)−∫cos⁡(log⁡x)dx]+C=x{sin⁡(log⁡x)+cos⁡(log⁡x)}−I+C[from Eq. (i) ]⇒I=x2{sin⁡(log⁡x)+cos⁡(log⁡x)}+C\begin{aligned} &I=\cos (\log x) \cdot x-\int[-\sin (\log x)] \cdot \frac{1}{x} \cdot x d x \\ &=x \cdot \cos (\log x)+\int \sin (\log x) \cdot 1 d x \\ &=x \cdot \cos (\log x)+\left[\sin (\log x) \cdot x-\int \cos (\log x) \cdot \frac{1}{x} \cdot x d x\right]+C \\ &=x \cdot(\log x)+\left[x \sin (\log x)-\int \cos (\log x) d x\right]+C \\ &=x\{\sin (\log x)+\cos (\log x)\}-I+C[f r o m \text{ Eq. (i) }] \\ &\Rightarrow I=\frac{x}{2}\{\sin (\log x)+\cos (\log x)\}+C \end{aligned})
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