How do you integrate #int x cos sqrtx dx # using integration by parts?

Question

How do you integrate #int x  cos  sqrtx   dx  # using integration by parts?

Isabella Ackerman · Accepted Answer

#2(sqrtxsin(sqrtx)(x-6)+3cos(sqrtx)(x-2))+C#

Before using integration by parts, let #t=sqrtx#. This implies that #t^2=x#. Differentiating this shows that #2tdt=dx#.

So:

#I=intxcos(sqrtx)dx=intt^2cos(t)(2tdt)=int2t^3cos(t)dt#

Now we should apply integration by parts. IBP takes the form #intudv=uv-intvdu#. So, for #int2t^3cos(t)dt#, let:

#{(u=2t^3,=>,du=6t^2dt),(dv=cos(t)dt,=>,v=sin(t)):}#

Then:

#I=uv-intvdu=2t^3sin(t)-int6t^2sin(t)dt#

For #int6t^2sin(t)dt#, use IBP again:

#{(u=6t^2,=>,du=12tdt),(dv=sin(t)dt,=>,v=-cos(t)):}#

Now:

#I=2t^3sin(t)-[-6t^2cos(t)+int12tcos(t)dt]#

#I=2t^3sin(t)+6t^2cos(t)-int12tcos(t)dt#

Reapplying IBP:

#{(u=12t,=>,du=12dt),(dv=cos(t)dt,=>,v=sin(t)):}#

#I=2t^3sin(t)+6t^2cos(t)-[12tsin(t)-int12sin(t)dt]#

#I=2t^3sin(t)+6t^2cos(t)-12tsin(t)+int12sin(t)dt#

Since #intsin(t)dt=-cos(t)#:

#I=2t^3sin(t)+6t^2cos(t)-12tsin(t)-12cos(t)#

Factoring:

#I=2(tsin(t)(t^2-6)+3cos(t)(t^2-2))#

Since #t=sqrtx#:

#I=2(sqrtxsin(sqrtx)(x-6)+3cos(sqrtx)(x-2))+C#

Aurora Alvarado · Answer

undefined To integrate $ \int x \cos(\sqrt{x}) \, dx $ using integration by parts, let $ u = x $ and $ dv = \cos(\sqrt{x}) \, dx $. Then, $ du = dx $ and $ v = 2\sqrt{x} \sin(\sqrt{x}) + 2 \cos(\sqrt{x}) $. Applying the integration by parts formula $ \int u \, dv = uv - \int v \, du $, we get:

$$
\begin{aligned}
\int x \cos(\sqrt{x}) \, dx &= x \left( 2\sqrt{x} \sin(\sqrt{x}) + 2 \cos(\sqrt{x}) \right) - \int \left( 2\sqrt{x} \sin(\sqrt{x}) + 2 \cos(\sqrt{x}) \right) \, dx \
&= 2x\sqrt{x} \sin(\sqrt{x}) + 2x\cos(\sqrt{x}) - 4\int \sqrt{x} \sin(\sqrt{x}) \, dx - 2\int \cos(\sqrt{x}) \, dx.
\end{aligned}
$$

Now, we can integrate $ \int \sqrt{x} \sin(\sqrt{x}) \, dx $ and $ \int \cos(\sqrt{x}) \, dx $ using integration by parts again or other suitable methods.