CBSE Class 12 · 2007 · Indefinite IntegralsLong Answer

Evaluate :

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

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

Solution

Put

t = cos 2 x

, so

d t = - 2 sin 2 x d x

, i.e.

sin 2 x d x = - \frac{1}{2} d t

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

Using

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

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

Restoring

t = cos 2 x

\frac{1}{2} lo g \frac{1 - cos 2 x}{2 - cos 2 x} + C

CBSE Class 12 Board · 2007 · Subjective