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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS1 Mark
Question
Choose the correct answer in Exercise:
$\int\frac{\cos2\text{x}}{(\sin\text{x}+\cos\text{x)}^{2}}$ is equal to
  1. $\frac{-1}{\sin\text{x}+\cos\text{x}}+\text{C}$
  2. $\log|\sin\text{x}+\cos\text{x}|+\text{C}$
  3. $\log|\sin\text{x}-\cos\text{x}|+\text{C}$
  4. $\frac{1}{(\sin\text{x}+\cos\text{x)}^{2}}+\text{C}$
✓

Answer

  1. $\log|\sin\text{x}+\cos\text{x}|+\text{C}$ 

$\text{Let I}=\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x)}^{2}}$

$\text{I}=\int\frac{(\cos^{2}\text{x}-\sin^{2}\text{x)}}{(\cos\text{x}+\sin\text{x)}^{2}}\text{dx}$

$=\int\frac{(\cos\text{x}+\sin\text{x})(\cos\text{x}-\sin\text{x)}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$

$=\int\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}+\sin\text{x}}\text{dx}$

$\text{Let}\cos\text{x}+\sin\text{x}=\text{t}\Rightarrow(\cos\text{x}-\sin\text{x)}\text{dx}=\text{dt}$

$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$

$=\log|\text{t}|+\text{C}$

$=\log|\cos\text{x}+\sin\text{x}|+\text{C}$

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