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CBSE Class 12 · 2007 · Indefinite IntegralsLong Answer

Evaluate :

\int \frac{s i n x}{( 1 - c o s x ) ( 2 - c o s x )} d x

Answer: $lo g ∣1 - cos x ∣ - lo g ∣2 - cos x ∣ + C$

Solution

Put

t = cos x

,

d t = - sin x d x

:

- \int \frac{d t}{( 1 - t ) ( 2 - t )} .

Partial fractions give

\frac{1}{( 1 - t ) ( 2 - t )} = \frac{1}{1 - t} - \frac{1}{2 - t}

, so

- \int [\frac{1}{1 - t} - \frac{1}{2 - t}] d t = lo g ∣1 - t ∣ - lo g ∣2 - t ∣ + C .

Restoring

t = cos x

,

lo g ∣1 - cos x ∣ - lo g ∣2 - cos x ∣ + C

CBSE Class 12 Board · 2007 · Subjective

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