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Integration of cos4x*cos5x*sin2x |
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Answer» 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 Learn More: INTEGRAL of 6x-5 root of 6-2x^2+xdx evaluate - Brainly.in integrate x³-1/x³+x dx - Brainly.in Integrate dx/ 3sin^2x+5cos^2x - Brainly.in |
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