How do you find the integral of $\int 48 (\frac{{sin (\sqrt{x})}^{3}}{\sqrt{x}}$ $int 48((sin(sqrt(x))^3)/(sqrt(x))$ ?

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Answer 1 · 2018-07-23T13:36:58

$4 cos (3 \sqrt{x}) - 36 cos (\sqrt{x}) + C$ $4\cos(3\sqrtx)-36\cos (\sqrtx)+C$

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

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

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

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

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

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

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

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

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