integration cos4x cos5x sin2x dx
Answers
Given : ∫cos4x cos5x sin2xdx
To Find : Integrate
Solution:
I = ∫cos4x cos5x sin2xdx
Simplify cos4x cos5x sin2x
= 2cos5x cos4x sin2x /2
2Cosacosb = Cos(a - b) + Cos(a + b)
= ( Cosx + cos 9x)sin2x/2
= Cosxsin2x/2 + cos9xsin2x/2
sin2x = 2sinx.cosx
= Cosxsinx.cosx + cos9xsin2x/2
= cos²xsinx + 2sin2xcos9x/4
2sinacosb = sin(a + b) + sin(a - b)
= cos²xsinx + {Sin11x + sin(-7x)}/4
= cos²xsinx + {Sin11x - sin(7x)}/4
= cos²xsinx + Sin11x/4 - Sin7x/4
I = ∫ cos²xsinx + Sin11x/4 - Sin7x/4) dx
= ∫(cos²xsinxdx +(1/4) ∫Sin11x .dx - (1/4) ∫Sin7x dx
∫(cos²xsinxdx
cosx = y , -sinxdx = dy
=> -∫y²dy = - y³/3 + c = -cos³x /3 + c
(1/4) ∫Sin11x .dx - (1/4) ∫Sin7x dx
= (1/4) -Cos11x/11 -(1/4)(-cos7x)/7 + c
I = -cos³x /3 - cos11x /44 + Cos7x/2x + C
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