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Integrate $\int t^2\sqrt{1-t}dt$

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Final Answer

$-\frac{2}{7}\sqrt{\left(1-t\right)^{7}}+\frac{4}{5}\sqrt{\left(1-t\right)^{5}}-\frac{2}{3}\sqrt{\left(1-t\right)^{3}}+C_0$
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Step-by-step Solution

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1

We can solve the integral $\int t^2\sqrt{1-t}dt$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $\sqrt{1-t}$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=\sqrt{1-t}$

Differentiate both sides of the equation $u=\sqrt{1-t}$

$du=\frac{d}{dt}\left(\sqrt{1-t}\right)$

Find the derivative

$\frac{d}{dt}\left(\sqrt{1-t}\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\frac{1}{2}\left(1-t\right)^{-\frac{1}{2}}\frac{d}{dt}\left(1-t\right)$

The derivative of a sum of two or more functions is the sum of the derivatives of each function

$\frac{1}{2}\left(1-t\right)^{-\frac{1}{2}}\left(\frac{d}{dt}\left(1\right)+\frac{d}{dt}\left(-t\right)\right)$

The derivative of the constant function ($1$) is equal to zero

$\frac{1}{2}\left(1-t\right)^{-\frac{1}{2}}\frac{d}{dt}\left(-t\right)$

The derivative of the linear function times a constant, is equal to the constant

$-\frac{1}{2}\left(1-t\right)^{-\frac{1}{2}}\frac{d}{dt}\left(t\right)$

The derivative of the linear function is equal to $1$

$-\frac{1}{2}\left(1-t\right)^{-\frac{1}{2}}$
2

Now, in order to rewrite $dt$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=-\frac{1}{2}\left(1-t\right)^{-\frac{1}{2}}dt$
3

Isolate $dt$ in the previous equation

$\frac{du}{-\frac{1}{2}\left(1-t\right)^{-\frac{1}{2}}}=dt$

Removing the variable's exponent raising both sides of the equation to the power of $2$

$\left(\sqrt{1-t}\right)^{\frac{1}{\frac{1}{2}}}=u^{\frac{1}{\frac{1}{2}}}$

Divide $1$ by $\frac{1}{2}$

$\left(\sqrt{1-t}\right)^{2}=u^{\frac{1}{\frac{1}{2}}}$

Simplify $\left(\sqrt{1-t}\right)^{2}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $\frac{1}{2}$ and $n$ equals $2$

$\left(1-t\right)^{\frac{1}{2}\cdot 2}$

Multiply $\frac{1}{2}$ times $2$

$1-t$

Multiply $\frac{1}{2}$ times $2$

$1-t=u^{\frac{1}{\frac{1}{2}}}$

Divide $1$ by $\frac{1}{2}$

$1-t=u^{2}$

We need to isolate the dependent variable $t$, we can do that by simultaneously subtracting $1$ from both sides of the equation

$-t=u^{2}-1$

Multiply both sides of the equation by $-1$

$t=-\left(u^{2}-1\right)$

Solve the product $-\left(u^{2}-1\right)$

$t=-u^{2}-1\cdot -1$

Multiply $-1$ times $-1$

$t=-u^{2}+1$
4

Rewriting $t$ in terms of $u$

$t=-u^{2}+1$

Rewriting $t$ in terms of $-u^{2}+1$

$\int\frac{\left(-u^{2}+1\right)^2u}{-\frac{1}{2}\left(1-\left(-u^{2}+1\right)\right)^{-\frac{1}{2}}}du$

Solve the product $-\left(-u^{2}+1\right)$

$\int\frac{\left(-u^{2}+1\right)^2u}{-\frac{1}{2}\left(1+u^{2}-1\right)^{-\frac{1}{2}}}du$

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\int\frac{\left(-u^{2}+1\right)^2u}{\frac{-1}{2\sqrt{1+u^{2}-1}}}du$

Subtract the values $1$ and $-1$

$\int\frac{\left(-u^{2}+1\right)^2u}{\frac{-1}{2\sqrt{u^{2}}}}du$

Cancel exponents $2$ and $\frac{1}{2}$

$\int\frac{\left(-u^{2}+1\right)^2u}{\frac{-1}{2u}}du$

Divide fractions $\frac{\left(-u^{2}+1\right)^2u}{\frac{-1}{2u}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

$\int2\left(-1\right)u^2\left(-u^{2}+1\right)^2du$

Multiply $2$ times $-1$

$\int-2u^2\left(-u^{2}+1\right)^2du$
5

Substituting $u$, $dt$ and $t$ in the integral and simplify

$\int-2u^2\left(-u^{2}+1\right)^2du$

Expand $\left(-u^{2}+1\right)^2$

$-2u^2\left(u^{4}-2u^{2}+1\right)$

Multiply the single term $-2u^2$ by each term of the polynomial $\left(u^{4}-2u^{2}+1\right)$

$-2u^{4}u^2+4u^{2}u^2-2u^2$

When multiplying exponents with same base we can add the exponents

$-2u^{6}+4u^{2}u^2-2u^2$

When multiplying two powers that have the same base ($u^{2}$), you can add the exponents

$-2u^{6}+4\left(u^{2}\right)^2-2u^2$

Simplify $\left(u^{2}\right)^2$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2$ and $n$ equals $2$

$-2u^{6}+4u^{2\cdot 2}-2u^2$

Simplify $\left(u^{2}\right)^2$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2$ and $n$ equals $2$

$-2u^2u^{2\cdot 2}+2\left(-1\right)\cdot 1u^{2}+1^2$

Multiply $2$ times $2$

$-2u^2u^{4}+2\left(-1\right)\cdot 1u^{2}+1^2$

Multiply $2$ times $2$

$-2u^{6}+4u^{4}-2u^2$
6

Rewrite the integrand $-2u^2\left(-u^{2}+1\right)^2$ in expanded form

$\int\left(-2u^{6}+4u^{4}-2u^2\right)du$
7

Expand the integral $\int\left(-2u^{6}+4u^{4}-2u^2\right)du$ into $3$ integrals using the sum rule for integrals, to then solve each integral separately

$\int-2u^{6}du+\int4u^{4}du+\int-2u^2du$

The integral of a function times a constant ($-2$) is equal to the constant times the integral of the function

$-2\int u^{6}du$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $6$

$-\frac{2}{7}u^{7}$

Replace $u$ with the value that we assigned to it in the beginning: $\sqrt{1-t}$

$-\frac{2}{7}\sqrt{\left(1-t\right)^{7}}$
8

The integral $\int-2u^{6}du$ results in: $-\frac{2}{7}\sqrt{\left(1-t\right)^{7}}$

$-\frac{2}{7}\sqrt{\left(1-t\right)^{7}}$

The integral of a function times a constant ($4$) is equal to the constant times the integral of the function

$4\int u^{4}du$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $4$

$\frac{4}{5}u^{5}$

Replace $u$ with the value that we assigned to it in the beginning: $\sqrt{1-t}$

$\frac{4}{5}\sqrt{\left(1-t\right)^{5}}$
9

The integral $\int4u^{4}du$ results in: $\frac{4}{5}\sqrt{\left(1-t\right)^{5}}$

$\frac{4}{5}\sqrt{\left(1-t\right)^{5}}$

The integral of a function times a constant ($-2$) is equal to the constant times the integral of the function

$-2\int u^2du$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $2$

$-\frac{2}{3}u^{3}$

Replace $u$ with the value that we assigned to it in the beginning: $\sqrt{1-t}$

$-\frac{2}{3}\sqrt{\left(1-t\right)^{3}}$
10

The integral $\int-2u^2du$ results in: $-\frac{2}{3}\sqrt{\left(1-t\right)^{3}}$

$-\frac{2}{3}\sqrt{\left(1-t\right)^{3}}$
11

Gather the results of all integrals

$-\frac{2}{7}\sqrt{\left(1-t\right)^{7}}+\frac{4}{5}\sqrt{\left(1-t\right)^{5}}-\frac{2}{3}\sqrt{\left(1-t\right)^{3}}$
12

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$-\frac{2}{7}\sqrt{\left(1-t\right)^{7}}+\frac{4}{5}\sqrt{\left(1-t\right)^{5}}-\frac{2}{3}\sqrt{\left(1-t\right)^{3}}+C_0$

Final Answer

$-\frac{2}{7}\sqrt{\left(1-t\right)^{7}}+\frac{4}{5}\sqrt{\left(1-t\right)^{5}}-\frac{2}{3}\sqrt{\left(1-t\right)^{3}}+C_0$

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Function Plot

Plotting: $-\frac{2}{7}\sqrt{\left(1-t\right)^{7}}+\frac{4}{5}\sqrt{\left(1-t\right)^{5}}-\frac{2}{3}\sqrt{\left(1-t\right)^{3}}+C_0$

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How to improve your answer:

Main Topic: Integrals with Radicals

Integrals with radicals are those integrals that contain a radical (square root, cubic, etc.) in the numerator or denominator of the integral.

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