Could you please explain how to integrate this equation: integral of x^3*sqrt(4x

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make a trigonometric substitution for x - x=2 sin t I = 8 sin^3 t sqrt(16 sin^2 t- 16 sin^4 t) = 32 sin^4 t sqrt(1-sin^2 t) = 32 sin^4 t cost then dx =2 cos t dt I = 16 sin^4t dt integral becomes 16 * integral ( sin^4t dt) integral of sin^4(x) =>[sin^2(t)]^2 =>[(1 -cos(2t)/2)]^2 =>1/4[1 + cos^2(2t) - 2cos(2t)] =>1/4[1 - 2cos(2t) + (1 + cos(4t)/2 ] =>(1/4) - (1/2)cos(2t) + (1/8) + (1/8)cos(4t) =>(3/8) - (1/2)cos(2t) + (1/8)cos(4t) so ?sin^4(t)dt = 3/8?dx - 1/2?cos(2t) dx + 1/8?cos(4t) dx ?sin^4(t)dt = (3/8)t - (1/2)sin(2t)*(1/2) + (1/8)sin(4t)*(1/4) + c ?sin^4(t)dt = (3/8)t - (1/4)sin(2t) + (1/32)sin(4t) + c again replace t = sin^-1(x/2) answer 16*{ (3/8)t - (1/4)sin(2t) + (1/32)sin(4t) + c} , where t = sin^-1(x/2)

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