Question
Evaluate the following integrals:
$\int\text{x}\Big(\frac{\sec2\text{x}-1}{\sec2\text{x}+1}\Big)\text{dx}$

Answer

Let $\text{I}=\int\text{x}\Big(\frac{\sec2\text{x}-1}{\sec2\text{x}+1}\Big)\text{dx}$
$=\int\text{x}\Big(\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\Big)\text{dx}$
$=\int\text{x}\Big(\frac{\sec^2\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\int\text{x}\tan^2\text{x dx}$
$=\int\text{x}(\sec^2\text{x}-1)\text{dx}$
$=\int\text{x}\sec^2\text{x dx}-\int\text{dx}$
$=\big[\text{x}\int\sec^2\text{x dx}-\int(1\int\sec^2\text{x dx})\text{dx}\big]-\frac{\text{x}^2}{2}$
$=\text{x}\tan\text{x}-\int\tan\text{x dx}-\frac{\text{x}^2}{2}$
$\text{I}=\text{x}\tan\text{x}-\log\sec\text{x}-\frac{\text{x}^2}{2}+\text{C}$

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