Indefinite Integrals — CBSE Class 12 · 2015 Board Question with Solution

Find

\int \frac{d x}{s i n x + s i n 2 x}

Answer: $\frac{1}{2} lo g ∣1 + cos x ∣ + \frac{1}{6} lo g ∣1 - cos x ∣ - \frac{2}{3} lo g ∣1 + 2 cos x ∣ + C$

Solution

Write

sin x + sin 2 x = sin x (1 + 2 cos x)

and multiply top and bottom by

sin x

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

Put

u = cos x

d u = - sin x d x

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

Resolving into partial fractions gives coefficients

\frac{1}{2}, \frac{1}{6}, - \frac{2}{3}

over

(1 + u), (1 - u), (1 + 2 u)

, so

\int [\frac{1}{2 ( 1 + u )} - \frac{1}{6 ( 1 - u )} - \frac{2}{3 ( 1 + 2 u )}] d u .

Integrating and restoring

u = cos x

\frac{1}{2} lo g ∣1 + cos x ∣ + \frac{1}{6} lo g ∣1 - cos x ∣ - \frac{2}{3} lo g ∣1 + 2 cos x ∣ + C

CBSE Class 12 Board · 2015 · Subjective