Sol. We have, (cos x + i sin x) 5 = cos 5x + i sin 5x

⇒ cos 5x + i sin 5x

⇒ ( 5 C 0 cos 5 x – 5 C 2 cos 3 x sin 2 x + 5 C 4 cos 4 x sin x)

+ i ( 5 C 1 cos 4 x sin x – 5 C 3 cos 2 x sin 3 x + 5 C 5 sin 5 x)

On equating imaginary parts, we get

sin 5x = 5 cos 4 x sin x –10 cos 2 x sin 3 x + sin 5 x

⇒ = 5 cos 4 x – 10 cos 2 x sin 2 x + sin 4 x

⇒ = 5 cos 4 x –10 cos 2 x (1 – cos 2 x) + (1 – cos 2 x) 2 ⇒ = 16 cos

4 x –12 cos 2 x + 1

Now,

16 cos 4 x – 12 cos 2 x + 1 = 0

⇒ cos 2 x = =

⇒ cos 2 x =

⇒ cos 2 x =

⇒ cos x = ±

⇒ cos x = , , – , –

⇒ cos x = cos , cos , cos , cos

⇒ x = , , ,

∴ 16 cos 4 x –12 cos 2 x + 1

= 16 (cos x – cos π /5) (cos x – cos 2 π /5) (cos x – cos 3 π /5) (cos x – cos 4 π /5)

= 16 (cos x – cos π /5) (cos x – cos 2 π /5) (cos x + cos 2 π /5) (cos x + cos π /5)

= 16(cos 2 x – cos 2 π /5) cos 2 x – cos 2 2 π /5)

= 16 (sin 2 π /5 – sin 2 x) (sin 2 2 π /5 – sin 2 x)

∴ =16 (sin 2 π /5– sin 2 x) (sin 2 2 π /5– sin 2 x)

⇒ =

= ×

= .

×

=

–

= –

= –

=

– sin

=

– sin

∴ dx

= sin dx

– sin dx

= sin

– sin + C

= sin log – sin log + C