Question
$\int\text{e}^{\text{x}}(1-\cot\text{x}+\cot^2\text{x})\text{dx}=$
  1. $\text{e}^{\text{x}}\cot\text{x}+\text{C}$
  2. $-\text{e}^{\text{x}}\cot\text{x}+\text{C}$
  3. $\text{e}^{\text{x}}\text{cosec x}+\text{C}$
  4. $-\text{e}^{\text{x}}\text{cosec x}+\text{C}$

Answer

  1. $-\text{e}^{\text{x}}\cot\text{x}+\text{C}$
Solution:
$\text{I}=\int\text{e}^{\text{x}}(1-\cot\text{x}+\cot^2\text{x})\text{dx}$
$\text{I}=\int\text{e}^\text{x}(1+\cot^2\text{x}-\cot\text{x})\text{dx}$
$\text{I}=\int\text{e}^{\text{x}}(\text{cosec}^2\text{x}-\cot\text{x})\text{dx}$
Here, $\text{f(x)}=-\cot\text{x}$
$\text{f}'(\text{x})=\text{cosec}^2\text{x}$
$\text{I}=-\text{e}^{\text{x}}\cot\text{x}+\text{C}$

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