MHB Integrating by Parts: Solving a Sin x Problem

[karush]

\\text{w8.4.13 Integration by Parts} nmh{2000}
$\displaystyle
I=\int \sin\left({\sqrt{x}}\right) \ d{t}
=2\sin\left({\sqrt{x}}\right)
-2\sqrt{x}\cos\left({\sqrt{x}}\right)+C$

$\begin{align}
\displaystyle
u& = {\sin\left({\sqrt{x}}\right)} &
dv&={1} \ dx \\ \\
du&={\frac{\cos\left({\sqrt{x}}\right)}{2\sqrt{x}}} dx&
v& ={x}
\end{align}$

$\displaystyle uv-\int v \ du $

$\displaystyle
I=
x\sin\left({\sqrt{x}}\right)
-\int \frac{x\cos\left(\sqrt{x}\right)}{2\sqrt{x}} \ dx $

$\text{now what ... or is there a better way to do this? } $

 

[Greg]

So far so good, I say! Now make the sub $w=\sqrt{x}$ and use integration by parts again. Watch out for sign errors and the like.

 

[karush]

$\displaystyle
w=\sqrt{x} \ \ \ x={w}^{2}$

$\displaystyle
I=
w^2 \sin\left({w}\right)
-\int
\frac{w^2 \cos\left(w\right)}
{2w } \ dw$
$\displaystyle
I=w^2 \sin\left({w}\right)
-\frac{1}{2}\int w\cos\left({w}\right) \ dw $

So far...

 

[MarkFL]

I think I would first let:

$$w=\sqrt{x}\,\therefore\,dx=2w\,dw$$

And we have:

$$I=2\int w\sin(w)\,dw$$

Now apply IBP:

$$u=w\,\therefore\,du=dw$$

$$dv=\sin(w)\,\therefore\,v=-\cos(w)$$

And we have:

$$I=2\left(-w\cos(w)+\int \cos(w)\,dw\right)=-2w\cos(w)+2\sin(w)+C$$

Now back-substitute for $w$:

$$I=2\sin(\sqrt{x})-2\sqrt{x}\cos(\sqrt{x})+C$$

 

[Greg]

Much quicker! At any rate, you may wish to try

$$\int\dfrac{x\cos(\sqrt x)}{2\sqrt x}\,dx$$

$$w=\sqrt x.\quad dw=\dfrac{1}{2\sqrt x}\,dx$$

$$\int w^2\cos w\,dw=w^2\sin w-2\int w\sin w\,dw$$

IBP again:

$$\int w^2\cos w\,dw=w^2\sin w-\left(-2w\cos w+2\int\cos w\,dw\right)$$

Can you continue?

 

[karush]

$$I=-2w\cos(w)+2\int \cos(w)\,dw=-2w\cos(w)+2\sin(w)+C$$
back-substitute for $w$ then

$$I=2\sin(\sqrt{x})-2\sqrt{x}\cos(\sqrt{x})+C$$

$\tiny\text{from Surf the Nations math study group}$