MHB Integrate by making substitution and by parts

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The discussion revolves around evaluating the integral of sin(√x) using substitution and integration by parts. Participants clarify that the "substitution" refers to a variable substitution, specifically u-substitution, where u = √x is proposed to eliminate the square root. The differential dx is expressed in terms of du, leading to a transformation of the integral into a more manageable form. The integration by parts is then applied, resulting in the final expression for the integral. The conversation emphasizes the importance of proper substitution to simplify the integration process.
find_the_fun
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I don't understand the question:

"First make a substitution and then use integration by parts to evaluate the integral"

[math]\int sin \sqrt{x} dx[/math]

What does it have in mind by "substitution"?
 
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Hi find_the_fun, (Wave)

By "substitution", they are referring to a variable substitution i.e u-substitution. What substitution would get rid of the square root? (Wondering)
 
Rido12 said:
Hi find_the_fun, (Wave)

By "substitution", they are referring to a variable substitution i.e u-substitution. What substitution would get rid of the square root? (Wondering)

So let [math]u=\sqrt{x}[/math] then [math]du=\frac{1}{2\sqrt{x}}dx[/math]
but none of that's in the original equation?
 
I would use the substitution $x=u^2$. This will eliminate the square root from the argument. (Nod)

I wouldn't square root both sides at this step, I would take the differential of both sides.
$dx =2u\cdot du$.

Can you proceed from here?

They are effectively the same, but taking the square root introduces more steps:

$$\sqrt{x}=u$$
$$\frac{1}{2\sqrt{x}}dx=du$$
$$dx=2\sqrt{x}\cdot du$$
Note that $\sqrt{x} = u$
$$dx=2u\cdot du$$
 
Last edited:
find_the_fun said:
I don't understand the question:

"First make a substitution and then use integration by parts to evaluate the integral"

[math]\int sin \sqrt{x} dx[/math]

What does it have in mind by "substitution"?

$\displaystyle \begin{align*} \int{ \sin{ \left( \sqrt{x} \right) } \,\mathrm{d}x } &= \int{ \frac{2\,\sqrt{x}\,\sin{ \left( \sqrt{x} \right) } }{2\,\sqrt{x}}\,\mathrm{d}x } \end{align*}$

Now let $\displaystyle \begin{align*} u = \sqrt{x} \implies \mathrm{d}u = \frac{1}{2\,\sqrt{x}}\,\mathrm{d}x \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{2\,\sqrt{x}\,\sin{ \left( \sqrt{x} \right) }}{2\,\sqrt{x}}\,\mathrm{d}x } &= \int{ 2\,u\sin{(u)}\,\mathrm{d}u} \\ &= -2\,u\cos{(u)} - \int{ -2\cos{(u)}\,\mathrm{d}u } \\ &= -2\,u\cos{(u)} + 2\int{ \cos{(u)}\,\mathrm{d}u } \\ &= -2\,u\cos{(u)} + 2\sin{(u)} + C \\ &= -2\,\sqrt{x}\,\cos{ \left( \sqrt{x} \right) } + 2\sin{ \left( \sqrt{x} \right) } + C \end{align*}$
 
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