# Antidifferentiation by Substitution

• MHB
• bunnypatotie
In summary, Dan is trying to solve a problem but is stuck because he doesn't know how to complete the square.
bunnypatotie
1.$\int x^2 e^{x^3} dx$
2. $\int sin(2x-3)dx$
3. $\int (\cfrac {3dx}{(x+2)\sqrt {x^2+4x+3}} )$
4. $\int (\cfrac {x^3}{(x^2 +4)^\cfrac {3}{2}} )dx$

Where are you having a problem? Show us what you've been able to do with these and we can help you much better.

-Dan

Once you get more experience, I think you will be able to see the substitution for problems easily!

For 1, let $u= x^3$. Then what is du? Do you see that "$x^2$" in the integral? What happens to that?

For 2, let $u= 2x- 3$. Then what is du?

For 3, let $u= x^2+ 4x+ 3$. Then what is du?
Be careful about the fact that "x+ 2" is in the denominator!

For 4, let $u= x^2+ 4$. What is du? Notice that $x^2dx= x^2(xdx)$ and $x^2= u- 4$.

HallsofIvy said:
Once you get more experience, I think you will be able to see the substitution for problems easily!

For 1, let $u= x^3$. Then what is du? Do you see that "$x^2$" in the integral? What happens to that?

For 2, let $u= 2x- 3$. Then what is du?

For 3, let $u= x^2+ 4x+ 3$. Then what is du?
Be careful about the fact that "x+ 2" is in the denominator!

For 4, let $u= x^2+ 4$. What is du? Notice that $x^2dx= x^2(xdx)$ and $x^2= u- 4$.
I'm already done with numbers 1, 3 and 4. I'm stuck with no.2 since I have no idea how to solve it

topsquark said:
Where are you having a problem? Show us what you've been able to do with these and we can help you much better.

-Dan
I'm done with items 1, 3 and 4. I'm stuck with item no. 2 since I don't have any idea how to solve this problem.

bunnypatotie said:
I'm already done with numbers 1, 3 and 4. I'm stuck with no.2 since I have no idea how to solve it
I meant no. 3 sorry. I'm done with nos. 1, 2 and 4 already. Haven't started with no. 3 yet because I don't have any idea how to really solve this problem

$\displaystyle 3\int \dfrac{dx}{(x+2)\sqrt{x^2+4x+4-1}} = 3\int \dfrac{dx}{(x+2)\sqrt{(x+2)^2-1}}$

let $u = x+2 \implies du = dx$ ...

$\displaystyle 3\int \dfrac{1}{u\sqrt{u^2-1}} \, du$

substitution again ...

$v = \sqrt{u^2-1}$

continue ...

skeeter said:
$\displaystyle 3\int \dfrac{dx}{(x+2)\sqrt{x^2+4x+4-1}} = 3\int \dfrac{dx}{(x+2)\sqrt{(x+2)^2-1}}$

let $u = x+2 \implies du = dx$ ...

$\displaystyle 3\int \dfrac{1}{u\sqrt{u^2-1}} \, du$

substitution again ...

$v = \sqrt{u^2-1}$

continue ...
Got it, tysm.

What skeeter did was "complete the square" $x^2+4x+ 3= x^3+ 4x+ 4- 4+ 3=(x^2+ 4x+ 4)- 1= (x- 2)^2- 1$.

$$latex$$

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## What is antidifferentiation by substitution?

Antidifferentiation by substitution is a method used in calculus to find the antiderivative of a function. It involves replacing a variable in the integrand with a new variable, and then solving the resulting integral.

## When is antidifferentiation by substitution used?

Antidifferentiation by substitution is used when the integrand contains a function within a function, such as sin(x2). It is also used when the integrand contains a variable raised to a power, such as x3.

## What is the general process for antidifferentiation by substitution?

The general process for antidifferentiation by substitution involves the following steps:

1. Identify the inner function and its derivative in the integrand.
2. Choose a new variable to replace the inner function.
3. Write the integral in terms of the new variable.
4. Find the antiderivative of the new integral.
5. Substitute the original variable back into the antiderivative.

## Can any function be solved using antidifferentiation by substitution?

No, not all functions can be solved using antidifferentiation by substitution. It is only applicable to certain types of functions, such as those mentioned in the answer to the second question. Other methods, such as integration by parts, may be needed for more complex functions.

## Are there any common mistakes to avoid when using antidifferentiation by substitution?

One common mistake to avoid is forgetting to substitute the original variable back into the antiderivative after solving the integral in terms of the new variable. Another mistake is choosing the wrong new variable, which can lead to incorrect solutions. It is important to carefully identify the inner function and choose a suitable new variable to ensure the correct solution.

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