Hi...I've been having problems with this question.
The Integral of [sin^3(sqrt(x)) ]/ (sqrt(x))
Should I set sqrt(x) = u and solve from there or should I bring the sqrt(x) in the denominator up?...I'm not even sure if either approach is correct.
Hi...I've been having problems with this question.
The Integral of [sin^3(sqrt(x)) ]/ (sqrt(x))
Should I set sqrt(x) = u and solve from there or should I bring the sqrt(x) in the denominator up?...I'm not even sure if either approach is correct.
$\displaystyle u = sin(\sqrt{x}) $
$\displaystyle \frac{du}{dx} = cos(\sqrt{x}) \times \frac{1}{2}x^{-\frac{1}{2}} $
$\displaystyle \frac{du}{dx} = \frac{cos(\sqrt{x})}{2\sqrt{x}} $
And the original integral can be written:
$\displaystyle \int \frac{sin(\sqrt{x})sin^2(\sqrt{x})}{\sqrt{x}}dx $
Better?
PS: This might actually be useless!
Oh ....is it possible to incorporate the reduction formula for sin into this?
The problem I'm having is with that rotten sqrt x in the denominator...even if I do bring it to the numerator, I've still got problems as to how to do it.
Is it possible to set sqrt x = v.
Then approach the problem as sin^3v / v ....I don't think it's possible...but I'm unsure.
I'm not sure on your approach to the problem?
Yes. It might clear things up.
By substituting $\displaystyle u = \sqrt{x}$, the integral becomes:
$\displaystyle 2 \int \sin^3{u}~du$
Take one factor of sin out and apply the Pythagorean identity ($\displaystyle \sin^2{\theta}+\cos^2{\theta} = 1$):
$\displaystyle 2 \int (1-\cos^2{u})\sin{u}~du$
Can you do it now?