Integrate by making substitution and by parts

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SUMMARY

The discussion focuses on evaluating the integral of sin(√x) using substitution and integration by parts. The recommended substitution is u = √x, which simplifies the integral by eliminating the square root. The differential dx is expressed as 2u du, leading to the transformed integral ∫ 2u sin(u) du. The final result is -2√x cos(√x) + 2sin(√x) + C, where C is the constant of integration.

PREREQUISITES
  • Understanding of u-substitution in calculus
  • Familiarity with integration by parts technique
  • Knowledge of trigonometric integrals
  • Ability to manipulate differentials and integrals
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  • Study the method of integration by parts in detail
  • Practice additional u-substitution problems
  • Explore trigonometric integrals involving sine and cosine functions
  • Learn about the properties of definite integrals and their applications
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Students and educators in calculus, mathematicians focusing on integral calculus, and anyone seeking to enhance their skills in solving integrals involving trigonometric functions.

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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