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.

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- Jan 12th 2009, 04:19 PMstorage1kIntegral of sin^3 (sqrt(x)) / (sqrt(x))
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. - Jan 12th 2009, 04:38 PMMush
$\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! - Jan 12th 2009, 05:13 PMstorage1k
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? - Jan 12th 2009, 05:24 PMChop Suey
Yes. (Nod) 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? - Jan 12th 2009, 05:37 PMstorage1k
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. - Jan 21st 2010, 04:48 PMChristian0412
Duh! lol Very helpful! Thanks!(Clapping)