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

Question

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

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Answer

Let $\text{I}=\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{dx}\ .....(\text{i})$
Let $\text{x}+\log\sin\text{x}=\text{t}$ then,
$\text{d}(\text{x}+\log\sin\text{x})=\text{dt}$
$\Rightarrow(1+\cot\text{x})\text{dx}=\text{dt}\Big[\because\frac{\text{d}}{\text{dx}}(\log\sin\text{x})=\cot\text{x}\Big]$
$\Rightarrow\text{dx}=\frac{\text{dt}}{1+\cot\text{x}}$
Putting $\text{x}+\log\sin\text{x}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{1+\cot\text{x}}$ In equation (i), we get,
$\text{I}=\int\frac{1+\cot\text{x}}{\text{t}}\times\frac{\text{dt}}{1+\cot\text{x}}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\text{x}+\log\sin\text{x}|+\text{C}$
$\therefore\text{I}=\log|\text{x}+\log\sin\text{x}|+\text{C}$

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