MHB Integration by parts curious question (chem's question at Yahoo Answers)

AI Thread Summary
The discussion centers on solving the integral of sin(x^1/2)/x^1/2 using integration by parts. The correct solution is identified as -2cos(x^1/2), with a detailed explanation provided on how to derive this result. The method involves recognizing the integral as an immediate form and applying integration by parts effectively. There is a debate about whether the solution relies on prior knowledge of the integral's value or if it can be derived independently. Ultimately, the conversation highlights the nuances of integration techniques and their applications in calculus.
Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
Here is the question:
Hi,

I am doing an integration by parts question but cannot work out how to get the solution. Any help would be greatly appreciated, cheers.

Integrate:

sin(x^1/2)/x^1/2

I know the solution is -2cos(x^1/2) but I do not know how to get to this.

Here is a link to the question:

Integration by parts question? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
Mathematics news on Phys.org
Hello chem,

We have an inmediate integral: $$\displaystyle\int\frac{\sin x^{1/2}}{x^{1/2}}dx=2\int\frac{\sin x^{1/2}}{2x^{1/2}}dx=2\int\sin x^{1/2}d(x^{1/2})=-2\cos x^{1/2}+C$$ Now, we can use the integration by parts method: $$\left \{ \begin{matrix}u=1\\dv=\frac{\sin x^{1/2}}{x^{1/2}}dx\end{matrix}\right.\Rightarrow \left \{ \begin{matrix}du=0dx\\v=-2\cos x^{1/2}\end{matrix}\right.\Rightarrow\\\int\frac{\sin x^{1/2}}{x^{1/2}}dx=1\cdot\left(-2\cos x^{1/2}\right)+\int 0\;dx=-2\cos x^{1/2}+C$$
 
Fernando Revilla said:
Hello chem,

We have an inmediate integral: $$\displaystyle\int\frac{\sin x^{1/2}}{x^{1/2}}dx=2\int\frac{\sin x^{1/2}}{2x^{1/2}}dx=2\int\sin x^{1/2}d(x^{1/2})=-2\cos x^{1/2}+C$$ Now, we can use the integration by parts method: $$\left \{ \begin{matrix}u=1\\dv=\frac{\sin x^{1/2}}{x^{1/2}}dx\end{matrix}\right.\Rightarrow \left \{ \begin{matrix}du=0dx\\v=-2\cos x^{1/2}\end{matrix}\right.\Rightarrow\\\int\frac{\sin x^{1/2}}{x^{1/2}}dx=1\cdot\left(-2\cos x^{1/2}\right)+\int 0\;dx=-2\cos x^{1/2}+C$$

Clearly You can proceed on this way only if You know a priori that...$$\int \frac{\sin \sqrt{x}}{\sqrt{x}}\ dx = -2\ \cos \sqrt{x} + c\ (1)$$... so that properly specking that is not an integration by parts. Very interesting is using (1) and integration by parts to arrive to an important result. Let's suppose to integrate by parts setting $u=\frac {\sin \sqrt{x}}{\sqrt{x}}$ and $v=1$... $$\int \frac{\sin \sqrt{x}}{\sqrt{x}}\ dx = \sqrt{x}\ \sin \sqrt{x} - \frac{1}{2}\ \int \cos \sqrt{x}\ dx + \frac{1}{2}\ \int \frac{\sin \sqrt{x}}{\sqrt{x}}\ dx\ (2)$$

... and from (1) and (2) You arrive to the result...

$$\int \cos \sqrt{x}\ dx = 2\ (\cos \sqrt{x} + \sqrt{x}\ \sin \sqrt{x}) + c\ (4)$$

In similar way You arrive to...

$$\int \sin \sqrt{x}\ dx = 2\ (\sin \sqrt{x} - \sqrt{x}\ \cos \sqrt{x}) + c\ (3)$$

Kind regards

$\chi$ $\sigma$
 
chisigma said:
Clearly You can proceed on this way only if You know a priori that... $$\int \frac{\sin \sqrt{x}}{\sqrt{x}}\ dx = -2\ \cos \sqrt{x} + c\ (1)$$

We don't suppose a priori the value of the given integral. We simply find $v=\displaystyle\int\frac{\sin x^{1/2}}{x^{1/2}}dx=\ldots=-2\cos x^{1/2}$ as an inmediate integral. It is irrelevant if we find it in the first or in the second line.

P.S. At any case, the title 'Integration by parts curious question' is meaningful. :)
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top