Question
Evaluate the following integrals:
$\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{dx}$
$\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{dx}$
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Answer
$\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}=\frac{\cos\text{x}-\sin\text{x}}{(\sin^2\text{x}+\cos^2\text{x})+2\sin\text{x}\cos\text{x}}$
$[\sin^2\text{x}+\cos^2\text{x}=1;\ \sin2\text{x}=2\sin\text{x}\cos\text{x}]$
$=\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}$
$\sin\text{x}+\cos\text{x}=\text{t}$
$\therefore(\sin\text{x}+\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{dx}=\int\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}\text{dx}$
$=\int\frac{\text{dt}}{\text{t}^2} $
$=\int\text{t}^{-2}\text{dt}$
$=-\text{t}^{-1}+\text{C}$
$=-\frac{1}{\text{t}}+\text{C}$
$=\frac{-1}{\sin\text{x}+\cos\text{x}}+\text{C}$
$[\sin^2\text{x}+\cos^2\text{x}=1;\ \sin2\text{x}=2\sin\text{x}\cos\text{x}]$
$=\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}$
$\sin\text{x}+\cos\text{x}=\text{t}$
$\therefore(\sin\text{x}+\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{dx}=\int\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}\text{dx}$
$=\int\frac{\text{dt}}{\text{t}^2} $
$=\int\text{t}^{-2}\text{dt}$
$=-\text{t}^{-1}+\text{C}$
$=-\frac{1}{\text{t}}+\text{C}$
$=\frac{-1}{\sin\text{x}+\cos\text{x}}+\text{C}$
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