# Integrate Sine and Square root Composite Function

#### Petrus

##### Well-known member
Hello MHB,
I got stuck on integrate this function

$$\displaystyle \int \frac{\sin^3(\sqrt{x})}{\sqrt{x}}dx$$
my first thinking was rewrite it as $$\displaystyle \int \frac{\sin^2(\sqrt{x})\sin(\sqrt{x})}{\sqrt{x}}dx$$
then use the identity $$\displaystyle \cos^2(x)+\sin^2(x)=1 \ \therefore \sin^2x=1- \cos^2(x)$$
$$\displaystyle \int \frac{(1-\cos^2(\sqrt{x}))\sin(\sqrt{x})}{\sqrt{x}}dx$$
subsitute $$\displaystyle u= \cos(x) \therefore du=- \sin(x) dx$$
then we get
$$\displaystyle - \int \frac{1-u^2}{ \cos^{-1}(u)}du$$
but that does not seem smart so my last ide is integrate by part, but I struggle on that part..

$$\displaystyle u= \sqrt{x}^{-1} \therefore du=\sqrt{x}^{-2}$$ and $$\displaystyle dv=\sin^3(\sqrt{x}) \therefore v=?$$

Regards,
$$\displaystyle |\pi\rangle$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Integrate 2

Use the substitution $\sqrt{x}=u$

#### Petrus

##### Well-known member
Re: Integrate 2

Use the substitution $\sqrt{x}=u$
Hello Zaid,
I don't see what is the point with that=S? can I substitute twice or ? I am kinda clueless Regards,
$$\displaystyle |\pi\rangle$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Integrate 2

Hello Zaid,
I don't see what is the point with that=S? can I substitute twice or ? I am kinda clueless Regards,
$$\displaystyle |\pi\rangle$$
One thing to realize that $$\displaystyle (\sqrt{x})'=\frac{1}{2\sqrt{x}}$$

So before we proceed , we make $$\displaystyle u=\sqrt{x}$$ substitution , which makes things easier since we are left with

$$\displaystyle 2\int \sin^3 (u)\,du$$ which you can integrate , right ?

I have got to get some sleep now , if you are still stuck someone is always around .

#### Petrus

##### Well-known member
Re: Integrate 2

One thing to realize that $$\displaystyle (\sqrt{x})'=\frac{1}{2\sqrt{x}}$$

So before we proceed , we make $$\displaystyle u=\sqrt{x}$$ substitution , which makes things easier since we are left with

$$\displaystyle 2\int \sin^3 (u)\,du$$ which you can integrate , right ?

I have got to get some sleep now , if you are still stuck someone is always around .
Hello Zaid,
Now it make alot sense! Thanks for taking your time and sleep well! I am also supposed to sleep but will do it soom =D

Regards,
$$\displaystyle |\pi\rangle$$