Question

1. evaluate the integral $\int \frac{\sec^{2}t\mathrm{d}t}{(\tan t - 1)^{2}}$.

Explanation:

Step1: Substitute $u = \tan t - 1$

Differentiate $u$ with respect to $t$: $du=\sec^{2}t\;dt$.
The integral $\int\frac{\sec^{2}t\;dt}{(\tan t - 1)^{2}}$ becomes $\int\frac{du}{u^{2}}$.

Step2: Integrate $\frac{1}{u^{2}}$

Recall the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$). For $n=-2$, we have $\int u^{-2}du=\frac{u^{-2 + 1}}{-2+1}+C$.
So, $\int u^{-2}du=-\frac{1}{u}+C$.

Step3: Substitute back $u=\tan t - 1$

Substituting $u = \tan t-1$ into $-\frac{1}{u}+C$, we get $-\frac{1}{\tan t - 1}+C=\frac{1}{1 - \tan t}+C$.

Answer:

$\frac{1}{1-\tan t}+C$