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Mathematics Question on integral

Integrate the rational function: cos x(1−sin x)(2−sin x)\frac {cos \ x}{(1-sin\ x)(2-sin\ x)}

Answer

cos x(1−sin x)(2−sin x)\frac {cos \ x}{(1-sin\ x)(2-sin\ x)}

Let sin x=t⇒cos x dx=dtLet \ sin \ x = t ⇒ cos \ x\ dx = dt

∴ ∫$$\frac {cos \ x}{(1-sin\ x)(2-sin\ x)}\ dx= ∫dt(1−t)(2−t)∫\frac {dt}{(1-t)(2-t)}

LetLet 1(1−t)(2−t)\frac {1}{(1-t)(2-t)} = A(1−t)+B(2−t)\frac {A}{(1-t)}+\frac {B}{(2-t)}

1=A(2−t)+B(1−t)1 = A(2-t)+B(1-t) ...(1) ...(1)

Substituting t=2 and then t=1 in equation (1),we obtainSubstituting\ t = 2 \ and \ then\ t = 1 \ in \ equation \ (1), we\ obtain

A=1 and B=−1A = 1\ and\ B = −1

∴ 1(1−t)(2−t)\frac {1}{(1-t)(2-t)} = 1(1−t)−1(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\ |1-t|+log\ |2-t|+C

= log ∣2−t1−t∣+Clog\ |\frac {2-t}{1-t}|+C

= log ∣2−sin x1−sin x∣+Clog\ |\frac {2-sin\ x}{1-sin\ x}|+C