# Can't Find a Correct Method to Integrate \int (t - 2)^2\sqrt{t}\,dt?

• KungPeng Zhou
KungPeng Zhou
Homework Statement
\int x^{2}\sqrt{2+x}dx
Relevant Equations
The Substitution Rule，
Table of Indefinite Integrals
When I encountereD this kind of question before.For example
\int x\sqrt{2+x^{2}}dx
We make the Substitution t=x^{2}+2，because its differential is dt=2xdx，so we get \int x\sqrt{2+x^{2}}=1/2\int\sqrt{t}dt，then we can get the answer easily
But the question，it seems that I can't use the way to solve the question.I can't find a correct commutation method.

Try putting ##$$## at head and ##$$## at tail of your math code to show it properly.

Last edited:
Sorry，but I can't understand you.Could you please tell how to show my math code properly?Now I just can use these math code.

KungPeng Zhou said:
Homework Statement: $\int x^{2}\sqrt{2+x}\,dx$
Relevant Equations: The Substitution Rule，
Table of Indefinite Integrals

When I encountereD this kind of question before.For example
$$\int x\sqrt{2+x^{2}}\,dx$$
We make the Substitution $t=x^{2}+2$，because its differential is $dt=2xdx[/tex]，so we get $$\int x\sqrt{2+x^{2}}\,dx =\frac12 \int\sqrt{t}\,dt,$$ then we can get the answer easily But the question，it seems that I can't use the way to solve the question.I can't find a correct commutation method. Did you consider [itex]t = x + 2$, $dx = dt$?

KungPeng Zhou said:
Sorry，but I can't understand you.Could you please tell how to show my math code properly?Now I just can use these math code.

Example for \frac{\pi}{2}
##\frac{\pi}{2}##
$$\frac{\pi}{2}$$

Why don't you try these two ways for Latex in your post ?

KungPeng Zhou said:
Sorry，but I can't understand you.Could you please tell how to show my math code properly?Now I just can use these math code.
Look at https://www.physicsforums.com/help/latexhelp/.
KungPeng Zhou said:
Homework Statement: ##\int x^{2}\sqrt{2+x}dx##
Relevant Equations: The Substitution Rule，
Table of Indefinite Integrals

When I encountereD this kind of question before.For example
##\int x\sqrt{2+x^{2}}dx##
We make the Substitution ##t=x^{2}+2##，because its differential is ##dt=2xdx##，so we get ##\int x\sqrt{2+x^{2}}=1/2\int\sqrt{t}dt##，then we can get the answer easily
But the question，it seems that I can't use the way to solve the question.I can't find a correct commutation method.
I'm not quite sure what you actually want to know. The LaTeX issue is addressed above and in the link.

The integral works as follows and is explained here:

pasmith said:
Did you consider $t = x + 2$, $dx = dt$?
Ithe seems that we still can't solve it with this way...

KungPeng Zhou said:
Ithe seems that we still can't solve it with this way...

You can't integrate $\int (t - 2)^2\sqrt{t}\,dt = \int t^{5/2} - 4t^{3/2} + 4t^{1/2}\,dt$?

KungPeng Zhou
pasmith said:
You can't integrate $\int (t - 2)^2\sqrt{t}\,dt = \int t^{5/2} - 4t^{3/2} + 4t^{1/2}\,dt$?
Yes，you are right.

Last edited by a moderator:

## What is the first step to integrate $$\int (t - 2)^2\sqrt{t}\,dt$$?

The first step is to expand the integrand. Rewrite $$(t - 2)^2$$ as $$t^2 - 4t + 4$$, then multiply by $$\sqrt{t}$$ to get $$\int (t^2 - 4t + 4)\sqrt{t}\,dt$$.

## How do you simplify the integrand $$\int (t^2 - 4t + 4)\sqrt{t}\,dt$$?

Distribute $$\sqrt{t}$$ to each term inside the parentheses: $$\int (t^2\sqrt{t} - 4t\sqrt{t} + 4\sqrt{t})\,dt$$, which simplifies to $$\int (t^{5/2} - 4t^{3/2} + 4t^{1/2})\,dt$$.

## What is the integral of each term in $$\int (t^{5/2} - 4t^{3/2} + 4t^{1/2})\,dt$$?

Integrate each term separately:$$\int t^{5/2}\,dt = \frac{2}{7}t^{7/2}$$,$$\int -4t^{3/2}\,dt = -4 \cdot \frac{2}{5}t^{5/2} = -\frac{8}{5}t^{5/2}$$,$$\int 4t^{1/2}\,dt = 4 \cdot \frac{2}{3}t^{3/2} = \frac{8}{3}t^{3/2}$$.

## What is the final form of the integral $$\int (t - 2)^2\sqrt{t}\,dt$$?

Combine the results from each term: $$\frac{2}{7}t^{7/2} - \frac{8}{5}t^{5/2} + \frac{8}{3}t^{3/2} + C$$, where $$C$$ is the constant of integration.

## Why is it important to include the constant of integration $$C$$ in the final answer?

The constant of integration $$C$$ is important because it represents the family of all antiderivatives. When integrating an indefinite integral, the solution includes all possible vertical shifts of the antiderivative function, which is captured by $$C$$.

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