Question
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\Big(\frac{\sin\text{x}\cos\text{x}-1}{\sin^2\text{x}}\Big)\text{dx}$
$\int\text{e}^{\text{x}}\Big(\frac{\sin\text{x}\cos\text{x}-1}{\sin^2\text{x}}\Big)\text{dx}$
✓
Answer
Let $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{\sin\text{x}\cos\text{x}-1}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\big[\cot\text{x}-\text{cosec}^2\text{x}\big]\text{dx}$
Here, $\text{f(x)}=\cot\text{x}$
$\Rightarrow\text{f}'\text{(x)}=-\text{cosec}^2\text{x}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\cot\text{x}=\text{t}$
Diff both sides w.r.t x
$\text{e}^{\text{x}}(\cot\text{x}-\text{cosec}^2\text{x})\text{dx = dt}$
$\therefore\text{I}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\cot\text{x + C}$
$=\int\text{e}^{\text{x}}\big[\cot\text{x}-\text{cosec}^2\text{x}\big]\text{dx}$
Here, $\text{f(x)}=\cot\text{x}$
$\Rightarrow\text{f}'\text{(x)}=-\text{cosec}^2\text{x}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\cot\text{x}=\text{t}$
Diff both sides w.r.t x
$\text{e}^{\text{x}}(\cot\text{x}-\text{cosec}^2\text{x})\text{dx = dt}$
$\therefore\text{I}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\cot\text{x + C}$
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