Integrate Sine and Square root Composite Function

[Petrus]

Hello MHB,
I got stuck on integrate this function

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

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

Regards,
$$|\pi\rangle$$

 

[alyafey22]

Re: Integrate 2

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

 

[Petrus]

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,
$$|\pi\rangle$$

 

[alyafey22]

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,
$$|\pi\rangle$$

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

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

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

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

 

[Petrus]

Re: Integrate 2

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

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

$$2\int \sin^3 (u)\,du $$ 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 a lot sense!:) Thanks for taking your time and sleep well! I am also supposed to sleep but will do it soom =D

Regards,
$$|\pi\rangle$$