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The integral:

there could be other ways. this is one.

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Another way:

we know that Sin 3x = 3 Sin x - 4 SIn³ x

=> Sin x Sin 3x = 3 Sin²x - 4 Sin⁴ x

=> 1/2 (Cos 2x - Cos 4x) = 3/2 (1-Cos2x) - 4 Sin⁴ x

=> Sin⁴ x = 1/8 [ 3 - 3 Cos 2x + Cos 4x - Cos 2x ]

=> Sin⁴ 2x = 1/8 [ 3 - 4 Cos 4x + Cos 8x ]

This is simpler, if you remember formula for sin 3x or sin 4x.

there could be other ways. this is one.

========================

Another way:

we know that Sin 3x = 3 Sin x - 4 SIn³ x

=> Sin x Sin 3x = 3 Sin²x - 4 Sin⁴ x

=> 1/2 (Cos 2x - Cos 4x) = 3/2 (1-Cos2x) - 4 Sin⁴ x

=> Sin⁴ x = 1/8 [ 3 - 3 Cos 2x + Cos 4x - Cos 2x ]

=> Sin⁴ 2x = 1/8 [ 3 - 4 Cos 4x + Cos 8x ]

This is simpler, if you remember formula for sin 3x or sin 4x.