CBSE Class 12 · 2006 · Indefinite IntegralsLong Answer

Evaluate :

\int \frac{d x}{1 + c o t x}

Answer: $\frac{1}{2} (x - lo g ∣ sin x + cos x ∣) + C$

Solution

Write

\frac{1}{1 + cot x} = \frac{sin x}{sin x + cos x}

. Express the numerator as

sin x = \frac{1}{2} [(sin x + cos x) - (cos x - sin x)],

where

cos x - sin x = \frac{d}{d x} (sin x + cos x)

. Hence

\int \frac{sin x}{sin x + cos x} d x = \frac{1}{2} \int d x - \frac{1}{2} \int \frac{cos x - sin x}{sin x + cos x} d x .

The second integral is

\frac{1}{2} lo g ∣ sin x + cos x ∣

, so

\frac{1}{2} (x - lo g ∣ sin x + cos x ∣) + C

CBSE Class 12 Board · 2006 · Subjective