Question
Evaluate the following integrals:$\int\frac{\sec^2\text{x}}{1-\tan^2\text{x}}\text{dx}$
✓
Answer
$\int\frac{\sec^2\text{x dx}}{1-\tan^2\text{x}}$
Let $\tan\text{x = t}$
$\Rightarrow\sec^2\text{x dx = dt}$
Now, $\int\frac{\sec^2\text{x dx}}{1-\tan^2\text{x}}$
$=\int\frac{\text{dt}}{1-\text{t}^2}$
$=\frac{1}{2}\log\bigg|\frac{1+\text{t}}{1-\text{t}}\bigg|+\text{C}$
$=\frac{1}{2}\log\bigg|\frac{1+\tan\text{x}}{1-\tan\text{x}}\bigg|+\text{C}$
Let $\tan\text{x = t}$
$\Rightarrow\sec^2\text{x dx = dt}$
Now, $\int\frac{\sec^2\text{x dx}}{1-\tan^2\text{x}}$
$=\int\frac{\text{dt}}{1-\text{t}^2}$
$=\frac{1}{2}\log\bigg|\frac{1+\text{t}}{1-\text{t}}\bigg|+\text{C}$
$=\frac{1}{2}\log\bigg|\frac{1+\tan\text{x}}{1-\tan\text{x}}\bigg|+\text{C}$
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