∫

sin(x−a)

sin(x+a)

dx
Let (x + a) = t ⇒ dx = dt
∴ I = ∫

sint−2a

sint

dt
= ∫

sintcos2a−costsin2a

sint

dt
= ∫ cos 2a - cot t sin 2a dt
= cos 2a t - sin 2a log |sin t| + C
= cos 2a x + a - sin 2a log |sin (x + a)| + C
OR
∫

5x−2

1+2x+3x2

dx
= 5 ∫

x− 2 5

1+2x+3x2

dx
=

5

6

∫

6x− 12 5

1+2x+3x2

dx
=

5

6

∫

6x+2− 12 5 −2

1+2x+3x2

dx
=

5

6

∫

6x+2− 22 5

1+2x+3x2

dx
=

5

6

∫

6x+2

1+2x+3x2

-

5

6

×

22

5

∫

1

3[(x+ 1 3 )2+ 2 9 ]

dx
=

5

6

log |1 + 2x + 3x2| -

11

9

∫

1

(x+ 1 3 )2+ 2 9

dx
=

5

6

log |1 + 2x + 3x2| -

11

9

×

3

√2

tan−1

(x+ 1 3 )

√2 3

+ C
=

5

6

log |1 + 2x + 3x2| -

11

3√2

× tan−1(

3x+1

√2

) + C