How do you find the integral of #int 48((sin(sqrt(x))^3)/(sqrt(x))#?

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Answer 1 · 2018-07-23T13:36:58

#4\cos(3\sqrtx)-36\cos (\sqrtx)+C#

Assuming there is a typo in the question, it can possibly taken as

#\int 48(\frac{(\sin(\sqrtx))^3}{\sqrtx})\ dx#

#=48\int\frac{\sin^3(\sqrtx)}{\sqrtx}\ dx#

Let #sqrtx=t\implies 1/{2\sqrtx}\ dx=dt\ \ or\ \ dx=2t dt#

#=48\int\frac{\sin^3t}{t}(2t\ dt)#

#=48\int \sin^3t\ dt#

#=48\int\frac{3\sint-\sin(3t)}{4}\ dt#

#=12(3\int \sin t\ dt-\int \sin(3t)\ dt) #

#=12(3(-\cos t)- \frac{-\cos(3t)}{3}) +C#

#=4\cos(3t)-36\cos t+C#

#=4\cos(3\sqrtx)-36\cos (\sqrtx)+C#