Question
Evaluate the following intregals:
$\int\frac{1}{1-\sin\text{x}+\cos\text{x}}\ \text{dx}$
$\int\frac{1}{1-\sin\text{x}+\cos\text{x}}\ \text{dx}$
✓
Answer
Let $\text{I}=\int\frac{1}{1-\sin\text{x}+\cos\text{x}}\ \text{dx}$
Putting $\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}},\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\text{I}=\int\frac{1}{1-\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}+\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}$
$=\int\frac{1+\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}-2\tan\frac{\text{x}}{2}+1-\tan^2\frac{\text{x}}{2}}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{2-2\tan\frac{\text{x}}{2}}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$=\frac{2}{2}\int\frac{\text{dt}}{1-\text{t}}$
$=-\log|1-\text{t}|+\text{C}$
$\text{I}=-\log\big|1-\tan\frac{\text{x}}{2}\big|+\text{C}$
Putting $\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}},\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\text{I}=\int\frac{1}{1-\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}+\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}$
$=\int\frac{1+\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}-2\tan\frac{\text{x}}{2}+1-\tan^2\frac{\text{x}}{2}}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{2-2\tan\frac{\text{x}}{2}}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$=\frac{2}{2}\int\frac{\text{dt}}{1-\text{t}}$
$=-\log|1-\text{t}|+\text{C}$
$\text{I}=-\log\big|1-\tan\frac{\text{x}}{2}\big|+\text{C}$
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