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

[Fernando Revilla]

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]

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]

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]

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. :)

 

[CellCoree]

Hello,

Thank you for reaching out for assistance with your integration by parts question. It seems like you have already attempted to solve the problem and have reached a certain point, but are unsure of how to proceed.

To integrate sin(x^1/2)/x^1/2, we will need to use the integration by parts method. This method involves breaking down the original function into two parts, u and dv, and then applying the following formula:

∫ u dv = uv - ∫ v du

In this case, let u = sin(x^1/2) and dv = 1/x^1/2. This means that du = (1/2)x^-1/2 cos(x^1/2) and v = 2x^1/2. Now, we can apply the formula:

∫ sin(x^1/2)/x^1/2 dx = sin(x^1/2) * 2x^1/2 - ∫ 2x^1/2 * (1/2)x^-1/2 cos(x^1/2) dx

Simplifying, we get:

∫ sin(x^1/2)/x^1/2 dx = 2sin(x^1/2)x^1/2 - ∫ cos(x^1/2) dx

Integrating cos(x^1/2) gives us -2sin(x^1/2), so the final solution is:

∫ sin(x^1/2)/x^1/2 dx = 2sin(x^1/2)x^1/2 + 2sin(x^1/2) + C

= 2sin(x^1/2)(x^1/2 + 1) + C

= -2cos(x^1/2) + C

I hope this helps you understand how to solve the problem. If you have any further questions, please do not hesitate to ask. Good luck with your studies!

Best regards,