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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
Choose the correct answer in Exercise : $\int\frac{\cos2\text{x}}{(\sin\text{x}+\cos\text{x)}^{2}}$ is equal to
  • A
    $\frac{-1}{\sin\text{x}+\cos\text{x}}+\text{C}$
  • ✓
    $\log|\sin\text{x}+\cos\text{x}|+\text{C}$
  • C
    $\log|\sin\text{x}-\cos\text{x}|+\text{C}$
  • D
    $\frac{1}{(\sin\text{x}+\cos\text{x)}^{2}}+\text{C}$
✓

Answer

Correct option: B.
$\log|\sin\text{x}+\cos\text{x}|+\text{C}$
Let $\text{ I}=\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x)}^{2}}$
$\text{I}=\int\frac{(\cos^{2}\text{x}-\sin^{2}\text{x)}}{(\cos\text{x}+\sin\text{x)}^{2}}\text{dx}$
$=\int\frac{(\cos\text{x}+\sin\text{x})(\cos\text{x}-\sin\text{x)}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}+\sin\text{x}}\text{dx}$
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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