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.

Originally Posted by storage1k

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?

Originally Posted by storage1k

Should I set sqrt(x) = u

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?

hehe! Thank you very much!

Worked it out quite nicely. Although I used
∫sin^n dx = -1/n cos x sin^(n-1) x + (n-1)/n ∫ sin^(n-2) xdx to go to it directly.

Duh! lol Very helpful! Thanks!