Topic : IntegrationQuestion : Solve by Integration using Trignometric Values∫Cos³ 2x Sin² 2x dxSolution :∫Cos³ 2x Sin² 2x dx
∫Cos² 2x Sin² 2x(Cos 2x) dx
Now Sin 2x = 2 Sin x Cos x
Sin 4x = 2 Sin 2x Cos 2x
Sin² 4x = 4 Sin² 2x Cos² 2x
Multiplying and dividing by 4
¼ ∫4 Sin² 2x Cos² 2x(Cos 2x) dx
¼ ∫ Sin² 4x Cos 2x dx
Now Cos 2x = 1 – 2 Sin² x
2 Sin² x = Cos 2x – 1
Sin² x = (Cos 2x – 1) / 2
Sin² 4x = (Cos 8x -1) / 2
So,
¼ ∫ [(Cos 8x -1) / 2] Cos 2x dx
1/8 ∫ (Cos 8x Cos 2x - Cos 2x) dx
1/8 [ ∫ (Cos 8x Cos 2x ) dx - ∫ Cos 2x dx ]
Now 2 Cos A Cos B = Cos (A+B) + Cos (A-B)
1/8 [ ∫ Cos 10x + Cos 6x dx - ∫ Cos 2x dx ]
1/8 {[ (Sin 10x)/10 + (Sin 6x)/6] – (Sin 2x)/2}
1/8 [(3 Sin 10x + 5 Sin 6x – 15 Sin 2x) / 30]
3 Sin 10x + 5 Sin 6x – 15 Sin 2x) / 240