Question
Evaluate the following integrals:
$\int\frac{\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$

Answer

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

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