Integrate by making substitution and by parts

In summary, to evaluate the integral $\int \sin \sqrt{x} dx$, we first make the substitution $u=\sqrt{x}$ and then use integration by parts to solve the integral. The final solution is $-2\sqrt{x}\cos(\sqrt{x}) + 2\sin(\sqrt{x}) + C$.
  • #1
find_the_fun
148
0
I don't understand the question:

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

\(\displaystyle \int sin \sqrt{x} dx\)

What does it have in mind by "substitution"?
 
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  • #2
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)
 
  • #3
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 \(\displaystyle u=\sqrt{x}\) then \(\displaystyle du=\frac{1}{2\sqrt{x}}dx\)
but none of that's in the original equation?
 
  • #4
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:
  • #5
find_the_fun said:
I don't understand the question:

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

\(\displaystyle \int sin \sqrt{x} dx\)

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

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

Related to Integrate by making substitution and by parts

1. What is the difference between integrating by substitution and by parts?

Integrating by substitution involves replacing a variable in the integrand with a new variable in order to simplify the integral. Integrating by parts involves using the product rule of differentiation to rewrite the integral in a different form.

2. How do I know when to use substitution or parts for integration?

In general, substitution is used when the integrand contains a single variable raised to a power, while parts is used when the integrand contains a product of two functions or a function raised to a power.

3. Can I use both substitution and parts in the same integration problem?

Yes, it is possible to use both methods in the same problem. In fact, in some cases it may be necessary to use both in order to solve the integral.

4. What is the purpose of using substitution and parts in integration?

The purpose of using these methods is to simplify the integral and make it easier to solve. In some cases, using substitution or parts can transform a difficult integral into one that is easier to evaluate.

5. Are there any tips for choosing the appropriate substitution or parts for integration?

When choosing a substitution, it is helpful to look for patterns in the integrand or to choose a variable that will cancel out terms in the integral. For parts, it is helpful to choose u and dv so that the resulting integral is simpler than the original one.

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