CBSE Class 12 · 2005 · Indefinite IntegralsVery Short Answer

Evaluate :

\int \frac{s i n 2 x}{a ^{2} s i n ^{2} x + b ^{2} c o s ^{2} x} d x

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

Solution

Put

t = a^{2} sin^{2} x + b^{2} cos^{2} x

so that differentiating gives

d t = (a^{2} \cdot 2 sin x cos x - b^{2} \cdot 2 sin x cos x) d x = (a^{2} - b^{2}) sin 2 x d x .

Hence

sin 2 x d x = \frac{d t}{a ^{2} - b ^{2}}

, and the integral becomes

\frac{1}{a ^{2} - b ^{2}} \int \frac{d t}{t} = \frac{1}{a ^{2} - b ^{2}} lo g ∣ t ∣ + C .

Restoring

t

\frac{1}{a ^{2} - b ^{2}} lo g a^{2} sin^{2} x + b^{2} cos^{2} x + C

CBSE Class 12 Board · 2005 · Subjective