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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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: $$\left \{ \begin{matrix}u=1\\dv=\frac{\sin x^{1/2}}{x^{1/2}}dx\end{matrix}\right.\Rightarrow \left \{ \begin{matrix}du=0dx\\v=-2\cos x^{1/2}\end{matrix}\right.\Rightarrow\\\int\frac{\sin x^{1/2}}{x^{1/2}}dx=1\cdot\left(-2\cos x^{1/2}\right)+\int 0\;dx=-2\cos x^{1/2}+C$$
 

chisigma

Well-known member
Feb 13, 2012
1,704
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

$\chi$ $\sigma$
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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. :)