It is known that, si A sin B =

1

2

cos A - B - cos A + B
∴ ∫ sin x sin 2x sin 3x dx = ∫ |sin x ×

1

2

cos 2x - 3x - cos 2x + 3x|
=

1

2

∫ sin x cos (-x) - sin x cos 5x dx
=

1

2

∫ sin x cos x - sin x cos 5x dx
=

1

2

∫

sin2x

2

dx -

1

2

∫ sin x cos 5x
=

1

4

|

−cos2x

2

| -

1

2

‌∫

1

2

sin x + 5x + sin x - 5x dx
=

cos2x

8

-

1

4

∫ (sin 6x + sin (-4x) dx
=

−cos2x

8

-

1

4

|

−cos6x

6

+

cos4x

4

| + C
=

−cos2x

8

-

1

8

|

−cos6x

3

+

cos4x

2

| + C
=

−6cos2x

48

-

1

8

|

−2cos6x+3cos4x

6

| + C
=

1

48

[2 cos 6x - 3 cos 4x - 6 cos 2x] + C
OR
Let

2

(1−x)(1+x2)

=

A

1−x

+

Bx+C

1+x2

2 = A (1+x2) + Bx + X (1 - x)
2 = A + Ax2 + Bx - Bx2 + C - Cx
Equating the coefficient of x2, x, and constant term, we obtain
A − B = 0
B − C = 0
A + C = 2
On solving these equations, we obtain
A = 1, B = 1, and C = 1
∴

2

(1−x)(1+x2)

=

1

1−x

+

x+1

1+x2

⇒ ∫

2

(1−x)(1+x2)

dx = ∫

1

1−x

dx + ∫

x

1+x2

dx + ∫

1

1+x2

dx
= - ∫

1

x−1

dx +

1

2

∫

2x

1+x2

dx + ∫

1

1+x2

dx
= - log |x - 1| +

1

2

log |1+x2| + tan−1 x + C