Put sin x = t
\(\frac {cos \ x}{(1-sin\ x)(2-sin\ x)}\)
\(Let \ sin \ x = t ⇒ cos \ x\ dx = dt\)
∴ \(∫\)\(\frac {cos \ x}{(1-sin\ x)(2-sin\ x)}\ dx\)= \(∫\frac {dt}{(1-t)(2-t)}\)
\(Let\) \(\frac {1}{(1-t)(2-t)}\) = \(\frac {A}{(1-t)}+\frac {B}{(2-t)}\)
\(1 = A(2-t)+B(1-t) \) \( ...(1)\)
\(Substituting\ t = 2 \ and \ then\ t = 1 \ in \ equation \ (1), we\ obtain\)
\(A = 1\ and\ B = −1\)
∴ \(\frac {1}{(1-t)(2-t)}\) = \(\frac {1}{(1-t)}-\frac {1}{(2-t)}\)
⇒ \(∫\)\(\frac {cos \ x}{(1-sin\ x)(2-sin\ x)}\ dx\) = \(∫\)\([\frac {1}{(1-t)}-\frac {1}{(2-t)}]\ dt\)
= \(-log\ |1-t|+log\ |2-t|+C\)
= \(log\ |\frac {2-t}{1-t}|+C\)
= \(log\ |\frac {2-sin\ x}{1-sin\ x}|+C\)
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Find the area of the region bounded by the curve y2=x and the lines x=1,x=4 and the x-axis
Find the area of the region bounded by y2=9x, x=2, x=4 and the x-axis in the first quadrant.
Find the area of the region bounded by x2=4y,y=2,y=4 and the x-axis in the first quadrant.
Find the area of the region bounded by the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\)
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,
