Integration: int [ (4 - r^2) (sqrt[4r^2 + 1]) r ] dr

[shaoen01]

Any help on this integration?

\(\displaystyle \int\, (4\, -\, r^{2})(\sqrt{4r^{2}\, +\, 1})r\, dr\)

 

[soroban]

Re: Integration help

Hello, shaoen01!

Integrate it by parts . . .

\(\displaystyle \L\int (4\,-\,r^{2})(\sqrt{4r^2\,+\,1})\,r\,dr\)

\(\displaystyle \begin{array}{ccccccc}u & = & 4\,-\,r^2 & \;\;\; & dv & = & (4r^2\,+\,1)^{\frac{1}{2}}\,r\,dr \\
du & = & -2r\,dr & & v & = & \frac{1}{12}(4r^2\,+\,1)^{\frac{3}{2}}
\end{array}\)

We have: \(\displaystyle \L\:\frac{1}{12}(4\,-\,r^2)(4r^2\,+\,1)^{\frac{3}{2}} \,+ \,\frac{1}{6}\int(4r^2\,+\,1)^{\frac{3}{2}}\,r\,dr\)

. . . . . \(\displaystyle \L=\;\frac{1}{12}(4\,-\,r^2)(4r^2\,+\,1)^{\frac{3}{2}} \,+\,\frac{1}{120}(4r^2\,+\,1)^{\frac{5}{2}}\,+\,C\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Technically, we're finished now,
. . but most books simplify the answer ... beyond all recognition!

I'll show you how they do it . . .

Factor: \(\displaystyle \L\:\frac{1}{120}\cdot(4r^2\,+\,1)^{\frac{3}{2}}\cdot\left[10(4\,-\,r^2)\,+\,(4r^2\,+\,1)\right]\,+\,C\)

. . . \(\displaystyle \L=\;\frac{1}{120}\cdot(4r^2\,+\,1)^{\frac{3}{2}}\cdot\left[40\,-\,10r^2 \,+\,4r^2\,+\,1\right]\,+\,C\)

. . . \(\displaystyle \L=\;\frac{1}{120}\cdot(4r^2\,+\,1)^{\frac{3}{2}}\cdot(41\,-\,6r^2)\,+\,C\;\;\) . . . There!

 

[shaoen01]

Re: Integration help

Hi Soroban,

How did you integrate \(\displaystyle \int{dv}\) to v? Sorry, usually i use another method for integration by parts, but i still kind of understand what you are doing. Thanks

 

[shaoen01]

Hi Soroban,

After much thought, i think i get it now! Thanks! But i am just wondering how you determine u or v is what value? What if you swapped both values, e.g. v to u and u to v. Will it make a difference? Thanks