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# Find the integral $\int x\cos\left(2x^2+3\right)dx$

## Step-by-step Solution

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###  Videos

$\frac{1}{4}\sin\left(2x^2+3\right)+C_0$
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##  Step-by-step Solution 

Problem to solve:

$\int x\cos\left(2x^2+3\right)dx$

Specify the solving method

1

We can solve the integral $\int x\cos\left(2x^2+3\right)dx$ 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 $2x^2+3$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=2x^2+3$

Differentiate both sides of the equation $u=2x^2+3$

$du=\frac{d}{dx}\left(2x^2+3\right)$

Find the derivative

$\frac{d}{dx}\left(2x^2+3\right)$

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

$\frac{d}{dx}\left(2x^2\right)+\frac{d}{dx}\left(3\right)$

The derivative of the constant function ($3$) is equal to zero

$\frac{d}{dx}\left(2x^2\right)$

The derivative of a function multiplied by a constant ($2$) is equal to the constant times the derivative of the function

$2\frac{d}{dx}\left(x^2\right)$

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

$4x$
2

Now, in order to rewrite $dx$ 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

$du=4xdx$
3

Isolate $dx$ in the previous equation

$\frac{du}{4x}=dx$

Simplify the fraction $\frac{x\cos\left(u\right)}{4x}$ by $x$

$\int\frac{\cos\left(u\right)}{4}du$
4

Substituting $u$ and $dx$ in the integral and simplify

$\int\frac{\cos\left(u\right)}{4}du$
5

Take the constant $\frac{1}{4}$ out of the integral

$\frac{1}{4}\int\cos\left(u\right)du$
6

We can solve the integral $\int\cos\left(u\right)du$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$

The derivative of the cosine of a function is equal to minus the sine of the function times the derivative of the function, in other words, if $f(x) = \cos(x)$, then $f'(x) = -\sin(x)\cdot D_x(x)$

$-\sin\left(u\right)$
7

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=\cos\left(u\right)}\\ \displaystyle{du=-\sin\left(u\right)du}\end{matrix}$
8

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=1du}\\ \displaystyle{\int dv=\int 1du}\end{matrix}$
9

Solve the integral

$v=\int1du$
10

The integral of a constant is equal to the constant times the integral's variable

$u$

The integral of a function times a constant ($-1$) is equal to the constant times the integral of the function

$\frac{1}{4}\left(u\cos\left(u\right)+1\int u\sin\left(u\right)du\right)$

Any expression multiplied by $1$ is equal to itself

$\frac{1}{4}\left(u\cos\left(u\right)+\int u\sin\left(u\right)du\right)$
11

Now replace the values of $u$, $du$ and $v$ in the last formula

$\frac{1}{4}\left(u\cos\left(u\right)+\int u\sin\left(u\right)du\right)$
12

Multiply the single term $\frac{1}{4}$ by each term of the polynomial $\left(u\cos\left(u\right)+\int u\sin\left(u\right)du\right)$

$\frac{1}{4}u\cos\left(u\right)+\frac{1}{4}\int u\sin\left(u\right)du$
13

We can solve the integral $\int u\sin\left(u\right)du$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$

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

$1$
14

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=u}\\ \displaystyle{du=du}\end{matrix}$
15

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\sin\left(u\right)du}\\ \displaystyle{\int dv=\int \sin\left(u\right)du}\end{matrix}$
16

Solve the integral

$v=\int\sin\left(u\right)du$
17

Apply the integral of the sine function: $\int\sin(x)dx=-\cos(x)$

$-\cos\left(u\right)$

The integral of a function times a constant ($-1$) is equal to the constant times the integral of the function

$\frac{1}{4}u\cos\left(u\right)+\frac{1}{4}\left(-u\cos\left(u\right)+1\int\cos\left(u\right)du\right)$

Any expression multiplied by $1$ is equal to itself

$\frac{1}{4}u\cos\left(u\right)+\frac{1}{4}\left(-u\cos\left(u\right)+\int\cos\left(u\right)du\right)$
18

Now replace the values of $u$, $du$ and $v$ in the last formula

$\frac{1}{4}u\cos\left(u\right)+\frac{1}{4}\left(-u\cos\left(u\right)+\int\cos\left(u\right)du\right)$

$\frac{1}{4}u\cos\left(u\right)- \left(\frac{1}{4}\right)u\cos\left(u\right)+\frac{1}{4}\int\cos\left(u\right)du$

Multiplying the fraction by $-1$

$\frac{1}{4}u\cos\left(u\right)-\frac{1}{4}u\cos\left(u\right)+\frac{1}{4}\int\cos\left(u\right)du$
19

Multiply the single term $\frac{1}{4}$ by each term of the polynomial $\left(-u\cos\left(u\right)+\int\cos\left(u\right)du\right)$

$\frac{1}{4}u\cos\left(u\right)-\frac{1}{4}u\cos\left(u\right)+\frac{1}{4}\int\cos\left(u\right)du$

Divide $1$ by $4$

$\frac{1}{4}u\cos\left(u\right)-\frac{1}{4}u\cos\left(u\right)+\frac{1}{4}\int\cos\left(u\right)du$

Multiplying the fraction by $u\cos\left(u\right)$

$\frac{1}{4}u\cos\left(u\right)+\frac{-u\cos\left(u\right)}{4}+\frac{1}{4}\int\cos\left(u\right)du$

Divide $1$ by $4$

$\frac{1}{4}u\cos\left(u\right)+\frac{-u\cos\left(u\right)}{4}+\frac{1}{4}\int\cos\left(u\right)du$

Combining like terms $\frac{1}{4}u\cos\left(u\right)$ and $\frac{-u\cos\left(u\right)}{4}$

$\left(\frac{1}{4}-\frac{1}{4}\right)u\cos\left(u\right)+\frac{1}{4}\int\cos\left(u\right)du$

Divide $-1$ by $4$

$\left(\frac{1}{4}-\frac{1}{4}\right)u\cos\left(u\right)+\frac{1}{4}\int\cos\left(u\right)du$

Subtract the values $\frac{1}{4}$ and $-\frac{1}{4}$

$0u+\frac{1}{4}\int\cos\left(u\right)du$

Any expression multiplied by $0$ is equal to $0$

$0+\frac{1}{4}\int\cos\left(u\right)du$

$x+0=x$, where $x$ is any expression

$\frac{1}{4}\int\cos\left(u\right)du$
20

Simplify the expression inside the integral

$\frac{1}{4}\int\cos\left(u\right)du$
21

Apply the integral of the cosine function: $\int\cos(x)dx=\sin(x)$

$\frac{1}{4}\sin\left(u\right)$

$\frac{1}{4}\sin\left(2x^2+3\right)$
22

Replace $u$ with the value that we assigned to it in the beginning: $2x^2+3$

$\frac{1}{4}\sin\left(2x^2+3\right)$
23

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

$\frac{1}{4}\sin\left(2x^2+3\right)+C_0$

$\frac{1}{4}\sin\left(2x^2+3\right)+C_0$

##  Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Solve integral of xcos(2x^2+3)dx using basic integralsSolve integral of xcos(2x^2+3)dx using u-substitutionSolve integral of xcos(2x^2+3)dx using tabular integration

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a
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u
v
w
x
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z
.
(◻)
+
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×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

### Main topic:

Integral Calculus

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