Evalute the following integrals: 1- 1+ dx - Vidyadip

Question

Evalute the following integrals:
$\int\frac{1-\cot\text{x}}{1+\cot\text{x}}\text{dx}$

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Answer

Let $\text{I}=\int\frac{1-\cot\text{x}}{1+\cot\text{x}}\text{dx}$ then,
$\text{I}=\int\frac{1-\frac{\cos\text{x}}{\sin\text{x}}}{1+\frac{\cos\text{x}}{\sin\text{x}}}\text{dx}$
$=\int\frac{\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}}}{\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}}}\text{dx}$
$\Rightarrow\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{dx}\ .....(\text{i})$
Let $\sin\text{x}+\cos\text{x}=\text{t},$ then,
$\text{d}(\sin\text{x}+\cos\text{x})=\text{dt}$
$\Rightarrow(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\Rightarrow-(\sin\text{x}-\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=-\frac{\text{dt}}{\sin\text{x}-\cos\text{x}}$
Putting $\sin\text{x}+\cos\text{x}=\text{t and dx}=-\frac{\text{dt}}{\sin\text{x}-\cos\text{x}}$ in equation (i), we het
$\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\text{t}}\times\frac{-\text{dt}}{\sin\text{x}-\cos\text{x}}$
$=\int\frac{-\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$=-\log|\sin\text{x}+\cos\text{x}|+\text{C}$

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