Put sin x = t
\(\frac {cos \ x}{(1-sin\ x)(2-sin\ x)}\)
\(Let \ sin \ x = t ⇒ cos \ x\ dx = dt\)
∴ \(∫\)\(\frac {cos \ x}{(1-sin\ x)(2-sin\ x)}\ dx\)= \(∫\frac {dt}{(1-t)(2-t)}\)
\(Let\) \(\frac {1}{(1-t)(2-t)}\) = \(\frac {A}{(1-t)}+\frac {B}{(2-t)}\)
\(1 = A(2-t)+B(1-t) \) \( ...(1)\)
\(Substituting\ t = 2 \ and \ then\ t = 1 \ in \ equation \ (1), we\ obtain\)
\(A = 1\ and\ B = −1\)
∴ \(\frac {1}{(1-t)(2-t)}\) = \(\frac {1}{(1-t)}-\frac {1}{(2-t)}\)
⇒ \(∫\)\(\frac {cos \ x}{(1-sin\ x)(2-sin\ x)}\ dx\) = \(∫\)\([\frac {1}{(1-t)}-\frac {1}{(2-t)}]\ dt\)
= \(-log\ |1-t|+log\ |2-t|+C\)
= \(log\ |\frac {2-t}{1-t}|+C\)
= \(log\ |\frac {2-sin\ x}{1-sin\ x}|+C\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]


The number of formulas used to decompose the given improper rational functions is given below. By using the given expressions, we can quickly write the integrand as a sum of proper rational functions.

For examples,
