Finding Integrals - Math

Card 0 of 252

Question

Evaluate:

$\dpi{120} \int_{0}^{\frac{\pi }{ 2}}\sin x \sqrt{\cos x }; ; dx$

Answer

Set $u = \cos x$ .

Then $\frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \cos x= -\sin x$ and $\sin x ; dx = - du$ .

Also, since $u = \cos x$ , the limits of integration change to $\cos \frac{\pi }{2} = 0$ and $\cos 0 = 1$ .

Substitute:

$\dpi{120} \int_{0}^{\frac{\pi }{ 2}}\sin x \sqrt{\cos x }; ; dx$

$= \int_{1}^{0} \left (- \sqrt{u} \right ) ; du$

$= \int_{1}^{0} \left (-u ^{\frac{1}{2}} \right ) ; du$

$= \int_{0}^{1} u ^{\frac{1}{2}} ; du$

$\dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \ & \end{matrix}\right|$ $\begin{matrix} 1\ \ 0 \end{matrix}$

$\dpi{150} = \left.\begin{matrix} \ \frac{ 2u \sqrt{u}} {3} & \ & \end{matrix}\right|$ $\begin{matrix} 1\ \ 0 \end{matrix}$

$=\frac{ 2\cdot 1\cdot \sqrt{1}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} = \frac{ 2} {3}$

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Question

Evaluate:

$\dpi{120}\int_{1}^{e^{ 3}}\frac{\sqrt{\ln x}}{x}; ; dx$

Answer

Set $u = \ln x$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and

$du = \frac{dx}{x}$

Also, since $u = \ln x$ , the limits of integration change to $\ln e^{3} = 3$ and $\ln 1= 0$ .

Substitute:

$\dpi{120}\int_{1}^{e^{ 3}}\frac{\sqrt{\ln x}}{x}; ; dx$

$\dpi{120} = \int_{0}^{3}\sqrt{u} ; du$

$= \int_{0}^{3} u ^{\frac{1}{2}} ; du$

$\dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \ & \end{matrix}\right|$ $\begin{matrix} 3\ \ 0 \end{matrix}$

$\dpi{150} = \left.\begin{matrix} \ \frac{ 2u \sqrt{u}} {3} & \ & \end{matrix}\right|$ $\begin{matrix} 3\ \ 0 \end{matrix}$

$=\frac{ 2\cdot 3\cdot \sqrt{3}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} =2\sqrt{3}$

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Question

Determine the indefinite integral:

$\dpi{120} \int\cos x \sqrt{\sin x }; ; dx$

Answer

Set $u = \sin x$ . Then

$\frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \sin x= \cos x$ .

and

$\cos x ; dx = du$

The integral becomes:

$\dpi{120} \int\cos x \sqrt{\sin x }; ; dx$

$\dpi{120} =\int \sqrt{u }; ; du$

$= \int u ^{\frac{1}{2}} ; du$

$\dpi{120} \dpi{150} = \frac{u^{\frac{1}{2}+1}}{{\frac{1}{2}+1}} + C$

$\dpi{150} = \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C$

$\dpi{120} =\frac{ 2u \sqrt{u}} {3} +C$

Substitute back:

$\dpi{120} =\frac{ 2 \sin x \cdot \sqrt{\sin x }} {3} +C$

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Question

Determine the indefinite integral:

$\dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}; ; dx$

Answer

$\log_{10} x = \frac{\ln x}{\ln 10}$ , so this can be rewritten as

$\dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}; ; dx$

$\dpi{120}=\int\frac{\sqrt{\ln x}}{x \sqrt{\ln 10} }; ; dx$

$\dpi{120}= \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }; ; dx$

Set $u = \ln x$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and

$du = \frac{dx}{x}$

Substitute:

$\dpi{120} \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }; ; dx$

$\dpi{120} =\frac{1}{\sqrt{\ln 10}} \int\sqrt{u} ; du$

$\dpi{120} =\frac{1}{\sqrt{\ln 10}} \int u ^{\frac{1}{2}} ; du$

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C \right )$

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2u \sqrt{u}} {3} + C \right )$

The outer factor can be absorbed into the constant, and we can substitute back:

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2 \ln x \cdot \sqrt{ \ln x}} {3} \right )+ C$

$= \frac{ 2 \sqrt{ \ln x} \cdot \ln x} {3 \cdot \sqrt{\ln 10}} + C$

$= \frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$

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Question

$\int_{4}^{9}\frac{\sqrt{x}}{1-x}dx=$

Answer

This integral will require a u-substitution.

Let $u=\sqrt{x}$ .

Then, differentiating both sides, $du=\frac{1}{2\sqrt{x}}dx$ .

We need to solve for dx in order to replace all x terms with u terms.

$dx=(2\sqrt{x})du$ .

$\int\frac{\sqrt{x}}{1-x}\cdot(2\sqrt{x})du=\int\frac{2xdu}{1-x}$

This is a little tricky because we stilll have x and u terms mixed together. We need to go back to our original substitution.

$u=\sqrt{x}\Rightarrow x=u^2$

$\int\frac{2xdu}{1-x}=\int\frac{2u^2 du}{1-u^2}$

Now we have an integral that looks more manageable. First, however, we can't forget about the bounds of the definite integral. We were asked to evaluate the integral from to . Because $u=\sqrt{x}$ , the bounds will change to $\sqrt{4}=2$ and $\sqrt{9}=3$ .

Essentially, we have made the following transformation:

$\int_{4}^{9}\frac{\sqrt{x}}{1-x}dx=\int_{2}^{3}\frac{2u^2}{1-u^2}du$ .

The latter integral is easier to evaluate.

$\int_{2}^{3}\frac{2u^2}{1-u^2}du=-2\int_{2}^{3}\frac{-u^2}{1-u^2}du$

$=-2\int_{2}^{3}\frac{1-u^2-1}{1-u^2}du=-2\int_{2}^{3}\left ( \frac{1-u^2}{1-u^2}+\frac{-1}{1-u^2} \right )du$

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

At this point, we can separate the integral into two smaller integrals.

$-2\int_{2}^{3}\left ( 1-\frac{1}{1-u^2}\right )du=-2\int_{2}^{3}1\cdot du+2\int_{2}^{3}\frac{1}{1-u^2}du$ .

The integral $-2\int_{2}^{3}1\cdot du$ evaluates to -2, so now we just need to worry about the other integral. This will require the use of partial fraction decomposition. We need to rewrite $\frac{1}{1-u^2}$ as the sum of two fractions.

$\frac{1}{1-u^2}=\frac{1}{(1-u)(1+u)}=\frac{A}{1-u}+\frac{B}{1+u}$

We need to solve for the values of A and B.

$\frac{A}{1-u}+\frac{B}{1+u}=\frac{1}{(1-u)(1+u)}$

This means that and . This is a relatively simple system of equations to solve, so I won't go into detail. The end result is that $A=B=\frac{1}2{}$ .

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

Let's now go back to the integral $2\int_{2}^{3}\frac{1}{1-u^2}du$ .

$2\int_{2}^{3}\frac{1}{1-u^2}du=2\int_{2}^{3}\left ( \frac{\frac{1}{2}}{1-u}+\frac{\frac{1}{2}}{1+u} \right )du$

Distribute the 2 to both integrals and separate it into two integrals.

$=\int_{2}^{3}\frac{1}{1-u}du + \int_{2}^{3}\frac{1}{1+u}du$

$=-\ln|1-u| ]_{2}^{3} + \ln|1+u|]_{2}^{3}$

$=-\ln|-2|+\ln|-1|+\ln4-\ln3$

$=-\ln2+\ln1+\ln4-\ln3$

$=\ln\frac{2}{3}$ .

Remember we need to add this value back to the value of $-2\int_{2}^{3}1\cdot du$ , which we already determined to be -2.

The final answer is $-2+\ln\frac{2}{3}$ .

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Question

$\int \frac{3}{x}=$

Answer

The integral of $\frac{1}{x}$ is $\ln \left | x \right |+C$ . The constant 3 is simply multiplied by the integral.

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Question

Evaluate the integral below:

$f(x)=\int \frac{dx}{x^2-16}$

Answer

In this case we have a rational function as $\frac{N(x)}{D(x)}$ , where

and

$D(x)=\frac{1}{x^2-16}$

can be written as a product of linear factors:

$\frac{1}{x^2-16}=\frac{1}{(x-4)(x+4)}\equiv \frac{A}{x-4}+\frac{B}{x+4}$

It is assumed that A and B are certain constants to be evaluated. Denominators can be cleared by multiplying both sides by (x - 4)(x + 4). So we get:

First we substitute x = -4 into the produced equation:

$1=-8B\Rightarrow B=-\frac{1}{8}$

Then we substitute x = 4 into the equation:

$1=8A\Rightarrow A=\frac{1}{8}$

Thus:

$\frac{1}{(x-4)(x+4)}=\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4}$

Hence:

$\int \frac{dx}{x^2-16}=\int(\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4})dx$

$=\frac{1}{8}ln\left | x-4 \right |-\frac{1}{8}ln\left |x+4 \right |+c$

$=\frac{1}{8}(ln\left | x-4 \right |-ln\left |x+4 \right |)+c$

$=\frac{1}{8}ln\left | \frac{x-4}{x+4}\right |+c$

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Question

$\int cos(x)\sin(x){\mathrm{d} x}=?$

Answer

To integrate $\cos(x)\sin(x)$ , we need to get the two equations in terms of each other. We are going to use "u-substitution" to create a new variable, , which will equal $\cos(x)$ .

Now, if $u=\cos(x)$ , then

$\frac{\mathrm{d}u }{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-sin(x)$

Multiply both sides by $\mathrm d {x}$ to get the more familiar:

${\mathrm{d} u}=-\sin(x)\mathrm{d}x$

Note that our $\mathrm d{u}=-\sin(x)$ , and our original equation was asking for a positive $\sin(x)$ .

That means if we want $\int\cos(x)\sin(x)$ in terms of , it looks like this:

$\int u\cdot (-\mathrm d{u}})$

Bring the negative sign to the outside:

$-\int u\mathrm d{u}$ .

We can use the power rule to find the integral of :

$-(\frac{1}{2}u^2+c)$

Since we said that $u=\cos(x)$ , we can plug that back into the equation to get our answer:

$-(\frac{1}{2}\cos^2(x)+c)$

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Question

What is the anti-derivative of ?

Answer

To find the indefinite integral of our expression, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

First we need to realize that . From there we can solve:

$\int2x{\mathrm{d} x}=\frac{2x^{1+1}}{1+1}+c$

When taking an integral, be sure to include a at the end of everything. stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being or instead of just .

$\int2x{\mathrm{d} x}=\frac{2x^{2}}{2}+c$

$\int2x{\mathrm{d} x}=x^2+c$

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Question

What is the indefinite integral of ?

Answer

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

Remember that, when taking the integral, we treat constants as that number times $x^{0}$ since anything to the zero power is . For example, treat as .

$\int x^3+2{\mathrm{d} x}=\frac{x^{3+1}}{3+1}+\frac{2x^{0+1}}{0+1}+c$

When taking an integral, be sure to include a at the end of everything. stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being or instead of just .

$\int x^3+2{\mathrm{d} x}=\frac{x^{4}}{4}+\frac{2x^{1}}{1}+c$

$\int x^3+2{\mathrm{d} x}=\frac{1}{4}x^4+2x+c$

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Question

What is the indefinite integral of ?

Answer

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

Remember that, when taking the integral, we treat constants as that number times $x^{0}$ , since anything to the zero power is . Treat as .

$\int x+18{\mathrm{d} x}=\frac{x^{1+1}}{1+1}+\frac{18x^{0+1}}{0+1}+c$

When taking an integral, be sure to include a . stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being or instead of just .

$\int x+18{\mathrm{d} x}=\frac{x^{2}}{2}+\frac{18x^{1}}{1}+c$

$\int x+18{\mathrm{d} x}=\frac{1}{2}x^2+18x+c$

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Question

What is the indefinite integral of $\sin(x)$ ?

Answer

$\sin(x)$ is a special function.

The indefinite integral is $\int \sin(x) {\mathrm{d} x}=-\cos(x)+c$ .

Even though it is a special function, we still need to include a . stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being or instead of just .

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Question

What is the indefinite integral of $\frac{1}{3}x^2$ ?

Answer

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

$\int \frac{1}{3}x^2{\mathrm{d} x}=\frac{1}{3}\frac{x^{2+1}}{2+1}+c$

When taking an integral, be sure to include a . stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being or instead of just .

$\int \frac{1}{3}x^2{\mathrm{d} x}=\frac{1}{3}\frac{x^{3}}{3}+c$

$\int \frac{1}{3}x^2{\mathrm{d} x}=\frac{1}{9}x^3+c$

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Question

What is the indefinite integral of ?

Answer

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

$\int x^2+5x{\mathrm{d} x}=\frac{x^{2+1}}{2+1}+\frac{5x^{1+1}}{1+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

$\int x^2+5x{\mathrm{d} x}=\frac{x^{3}}{3}+\frac{5x^{2}}{2}+c$

$\int x^2+5x{\mathrm{d} x}=\frac{1}{3}x^3+\frac{5}{2}x^2+c$

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Question

What is the indefinite integral of ?

Answer

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

$\int 5x+8{\mathrm{d} x}=\frac{5x^{1+1}}{1+1}+\frac{8x^{0+1}}{0+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

$\int 5x+8{\mathrm{d} x}=\frac{5x^{2}}{2}+\frac{8x^{1}}{1}+c$

$\int 5x+8{\mathrm{d} x}=\frac{5}{2}x^2+8x +c$

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Question

What is the indefinite integral of ?

Answer

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent.

We're going to treat as , as anything to the zero power is one.

For this problem, that would look like:

$\int x^3+2x+5{\mathrm{d} x}=\frac{x^{3+1}}{3+1}+\frac{2x^{1+1}}{1+1}+\frac{5x^{0+1}}{0+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

$\int x^3+2x+5{\mathrm{d} x}=\frac{x^{4}}{4}+\frac{2x^{2}}{2}+\frac{5x^{1}}{1}+c$

$\int x^3+2x+5{\mathrm{d} x}=\frac{1}{4}x^4+x^2+5x+c$

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Question

What is the indefinite integral of ?

Answer

To solve this problem, we can use the anti-power rule or reverse power rule. We raise the exponent on the variables by one and divide by the new exponent.

For this problem, we'll treat as since anything to the zero power is one.

$\int 6x^2+8{\mathrm{d} x}=\frac{6x^{2+1}}{2+1}+\frac{8x^{0+1}}{0+1}+c$

Since the derivative of any constant is , when we take the indefinite integral, we add a to compensate for any constant that might be there.

From here we can simplify.

$\int 6x^2+8{\mathrm{d} x}=\frac{6x^{3}}{3}+\frac{8x^{1}}{1}+c$

$\int 6x^2+8{\mathrm{d} x}=2x^3+8x+c$

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Question

What is the indefinite integral of ?

Answer

To solve this problem, we can use the anti-power rule or reverse power rule. We raise the exponent on the variables by one and divide by the new exponent.

For this problem, we'll treat as since anything to the zero power is one.

$\int x+1{\mathrm{d} x}=\frac{x^{1+1}}{1+1}+\frac{1x^{0+1}}{0+1}+c$

Since the derivative of any constant is , when we take the indefinite integral, we add a to compensate for any constant that might be there.

From here we can simplify.

$\int x+1{\mathrm{d} x}=\frac{x^{2}}{2}+\frac{x^{1}}{1}+c$

$\int x+1{\mathrm{d} x}=\frac{1}{2}x^2+x+c$

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Question

What is the indefinite integral of ?

Answer

To solve this problem, we can use the anti-power rule or reverse power rule. We raise the exponent on the variables by one and divide by the new exponent.

For this problem, we'll treat as since anything to the zero power is one.

$\int 12x^2+13x+4{\mathrm{d} x}=\frac{12x^{2+1}}{2+1}+\frac{13x^{1+1}}{1+1}+\frac{4x^{0+1}}{0+1}+c$

Since the derivative of any constant is , when we take the indefinite integral, we add a to compensate for any constant that might be there.

From here we can simplify.

$\int 12x^2+13x+4{\mathrm{d} x}=\frac{12x^{3}}{3}+\frac{13x^{2}}{2}+\frac{4x^{1}}{1}+c$

$\int 12x^2+13x+4{\mathrm{d} x}=4x^3+\frac{13}{2}x^2+4x+c$

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Question

What is the indefinite integral of $2x^4+\frac{1}{2}x$ ?

Answer

To find the indefinite integral, we can use the reverse power rule: we raise the exponent by one and then divide by our new exponent.

$\int 2x^4+\frac{1}{2}x{\mathrm{d} x}=\frac{2x^{4+1}}{4+1}+\frac{\frac{1}{2}x^{1+1}}{1+1}+c$

Remember when taking the indefinite integral to include a to cover any potential constants.

Simplify.

$\int 2x^4+\frac{1}{2}x{\mathrm{d} x}=\frac{2x^{5}}{5}+\frac{\frac{1}{2}x^{2}}{2}+c$

$\int 2x^4+\frac{1}{2}x{\mathrm{d} x}=\frac{2}{5}x^5+\frac{1}{4}x^{2}+c$

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