Evaluate the following integrals: 1 1- dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1}{1-\sin\text{x}}\text{dx}$

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Answer

$\int\frac{1}{1-\sin\text{x}}\text{dx}$$=\int\frac{(1+\sin\text{x})}{(1-\sin\text{x})\times(1+\sin\text{x})}\text{dx}$
$=\Big(\frac{1+\sin\text{x}}{1-\sin^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1+\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1}{\cos^2\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}\times\frac{1}{\cos\text{x}}\Big)\text{dx}$
$=\int(\sec^2\text{x}+\sec\text{x}\tan\text{x})\text{dx}$
$=\tan\text{x}+\sec\text{x}+\text{C}$

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