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CBSE Class 12 · 2013 · Indefinite IntegralsLong Answer

Evaluate :

\int \frac{s i n ( x - a )}{s i n ( x + a )} d x

Answer: $(x + a) cos 2 a - sin 2 a lo g ∣ sin (x + a) ∣ + C$

Solution

Put

t = x + a

, so

d x = d t

and

x - a = t - 2 a

. Expanding,

\frac{sin ( t - 2 a )}{sin t} = cos 2 a - sin 2 a cot t .

Integrating,

\int (cos 2 a - sin 2 a cot t) d t = t cos 2 a - sin 2 a lo g ∣ sin t ∣ + C .

Restoring

t = x + a

,

(x + a) cos 2 a - sin 2 a lo g ∣ sin (x + a) ∣ + C

CBSE Class 12 Board · 2013 · Subjective

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