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

Step-by-step Solution

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Final Answer

$\frac{1}{4}x\left(2x+\sin\left(2x\right)\right)-\frac{1}{4}x^2+\frac{1}{8}\cos\left(2x\right)+C_0$
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Step-by-step Solution

Specify the solving method

1

Apply the trigonometric identity: $\cos\left(\theta \right)^2$$=\frac{1+\cos\left(2\theta \right)}{2}$

$\int x\frac{1+\cos\left(2x\right)}{2}dx$
2

Multiplying the fraction by $x$

$\int\frac{x\left(1+\cos\left(2x\right)\right)}{2}dx$
3

Take the constant $\frac{1}{2}$ out of the integral

$\frac{1}{2}\int x\left(1+\cos\left(2x\right)\right)dx$
4

Divide $1$ by $2$

$\frac{1}{2}\int x\left(1+\cos\left(2x\right)\right)dx$
5

We can solve the integral $\int x\left(1+\cos\left(2x\right)\right)dx$ 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 $2x$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=2x$

Differentiate both sides of the equation $u=2x$

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

Find the derivative

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

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

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

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

$2$
6

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

$du=2dx$
7

Isolate $dx$ in the previous equation

$du=2dx$

Divide both sides of the equation by $2$

$\frac{2x}{2}=\frac{u}{2}$

Simplifying the quotients

$x=\frac{u}{2}$
8

Rewriting $x$ in terms of $u$

$x=\frac{u}{2}$

Rewriting $x$ in terms of $\frac{u}{2}$

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

Multiplying the fraction by $1+\cos\left(u\right)$

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

Solve the product $u\left(1+\cos\left(u\right)\right)$

$\int\frac{\frac{u+u\cos\left(u\right)}{2}}{2}du$

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

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

Factor the polynomial $u+u\cos\left(u\right)$ by it's greatest common factor (GCF): $u$

$\int\frac{u\left(1+\cos\left(u\right)\right)}{4}du$
9

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

$\frac{1}{2}\int\frac{u\left(1+\cos\left(u\right)\right)}{4}du$
10

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

$\frac{1}{2}\cdot \left(\frac{1}{4}\right)\int u\left(1+\cos\left(u\right)\right)du$

Divide $1$ by $4$

$\frac{1}{2}\cdot \frac{1}{4}\int u\left(1+\cos\left(u\right)\right)du$

Multiply $\frac{1}{2}$ times $\frac{1}{4}$

$\frac{1}{8}\int u\left(1+\cos\left(u\right)\right)du$
11

Simplify the expression inside the integral

$\frac{1}{8}\int u\left(1+\cos\left(u\right)\right)du$
12

We can solve the integral $\int u\left(1+\cos\left(u\right)\right)du$ 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 $1$

$1$
13

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=u}\\ \displaystyle{du=du}\end{matrix}$
14

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\left(1+\cos\left(u\right)\right)du}\\ \displaystyle{\int dv=\int \left(1+\cos\left(u\right)\right)du}\end{matrix}$
15

Solve the integral

$v=\int\left(1+\cos\left(u\right)\right)du$
16

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

$u+\int\cos\left(u\right)du$
17

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

$u+\sin\left(u\right)$

Expand the integral $\int\left(u+\sin\left(u\right)\right)du$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

$\frac{1}{8}\left(u\left(u+\sin\left(u\right)\right)-\int udu-\int\sin\left(u\right)du\right)$
18

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

$\frac{1}{8}\left(u\left(u+\sin\left(u\right)\right)-\int udu-\int\sin\left(u\right)du\right)$
19

Solve the product $\frac{1}{8}\left(u\left(u+\sin\left(u\right)\right)-\int udu-\int\sin\left(u\right)du\right)$

$\frac{1}{8}u\left(u+\sin\left(u\right)\right)+\frac{1}{8}\left(-\int udu-\int\sin\left(u\right)du\right)$

$\frac{1}{8}\cdot 2x\left(2x+\sin\left(2x\right)\right)+\frac{1}{8}\left(-\int udu-\int\sin\left(u\right)du\right)$

Multiply $\frac{1}{8}$ times $2$

$\frac{1}{4}x\left(2x+\sin\left(2x\right)\right)+\frac{1}{8}\left(-\int udu-\int\sin\left(u\right)du\right)$
20

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

$\frac{1}{4}x\left(2x+\sin\left(2x\right)\right)+\frac{1}{8}\left(-\int udu-\int\sin\left(u\right)du\right)$
21

Solve the product $\frac{1}{8}\left(-\int udu-\int\sin\left(u\right)du\right)$

$\frac{1}{4}x\left(2x+\sin\left(2x\right)\right)-\frac{1}{8}\int udu-\frac{1}{8}\int\sin\left(u\right)du$

Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, in this case $n=1$

$-\frac{1}{16}u^2$

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

$-\frac{1}{16}\left(2x\right)^2$

The power of a product is equal to the product of it's factors raised to the same power

$-\frac{1}{4}x^2$
22

The integral $-\frac{1}{8}\int udu$ results in: $-\frac{1}{4}x^2$

$-\frac{1}{4}x^2$

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

$\frac{1}{8}\cos\left(u\right)$

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

$\frac{1}{8}\cos\left(2x\right)$
23

The integral $-\frac{1}{8}\int\sin\left(u\right)du$ results in: $\frac{1}{8}\cos\left(2x\right)$

$\frac{1}{8}\cos\left(2x\right)$
24

Gather the results of all integrals

$\frac{1}{4}x\left(2x+\sin\left(2x\right)\right)-\frac{1}{4}x^2+\frac{1}{8}\cos\left(2x\right)$
25

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

$\frac{1}{4}x\left(2x+\sin\left(2x\right)\right)-\frac{1}{4}x^2+\frac{1}{8}\cos\left(2x\right)+C_0$

Final Answer

$\frac{1}{4}x\left(2x+\sin\left(2x\right)\right)-\frac{1}{4}x^2+\frac{1}{8}\cos\left(2x\right)+C_0$

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 xcosx^2dx using basic integralsSolve integral of xcosx^2dx using u-substitutionSolve integral of xcosx^2dx using integration by partsSolve integral of xcosx^2dx using tabular integration

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Function Plot

Plotting: $\frac{1}{4}x\left(2x+\sin\left(2x\right)\right)-\frac{1}{4}x^2+\frac{1}{8}\cos\left(2x\right)+C_0$

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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

How to improve your answer:

Main Topic: Integral Calculus

Integration assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data.

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