Question

what is the value of integration of sin cube of root x?

Answer 1

u = sin(\sqrt{x})

\frac{du}{dx} = cos(\sqrt{x}) \times \frac{1}{2}x^{-\frac{1}{2}}


\frac{du}{dx} = \frac{cos(\sqrt{x})}{2\sqrt{x}}

And the original integral can be written:

\int \frac{sin(\sqrt{x})sin^2(\sqrt{x})}{\sqrt{x}}dx

Answer 2

By substituting u = \sqrt{x}, the integral becomes:
2 \int \sin^3{u}~du

Take one factor of sin out and apply the Pythagorean identity ( \sin^2{\theta}+\cos^2{\theta} = 1):
2 \int (1-\cos^2{u})\sin{u}~du

Can you do it now?