Integral of sin^3 (sqrt(x)) / (sqrt(x))

**storage1k** · January 12th 2009, 04:19 PM

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.

**Mush** · January 12th 2009, 04:38 PM

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.

$u = sin(\sqrt{x})$

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

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

And the original integral can be written:

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

Better?

PS: This might actually be useless!

**storage1k** · January 12th 2009, 05:13 PM

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?

**Chop Suey** · January 12th 2009, 05:24 PM

Originally Posted by storage1k

Should I set sqrt(x) = u

Yes.

It might clear things up.

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?

**storage1k** · January 12th 2009, 05:37 PM

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.

**Christian0412** · January 21st 2010, 04:48 PM

Duh! lol Very helpful! Thanks!

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