Step-by-step Solution

Solve the trigonometric integral $\int\frac{1}{2\sin\left(x\right)\cos\left(x\right)}dx$

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Final Answer

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

Step-by-step Solution

Problem to solve:

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

Solving method

1

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

$u=\sin\left(x\right)$

Differentiate both sides of the equation $u=\sin\left(x\right)$

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

Find the derivative

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

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

$\cos\left(x\right)$
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

$du=\cos\left(x\right)dx$
3

Isolate $dx$ in the previous equation

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

Take the inverse of $\sin\left(x\right)$ on both sides

$\arcsin\left(\sin\left(x\right)\right)=\arcsin\left(u\right)$

Apply the formula: $\arcsin\left(\sin\left(x\right)\right)$$=x$

$x=\arcsin\left(u\right)$
4

Rewriting $x$ in terms of $u$

$x=\arcsin\left(u\right)$

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

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

Rewriting $x$ in terms of $\arcsin\left(u\right)$

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

Simplify $\cos\left(\arcsin\left(u\right)\right)$ as $\sqrt{1-u^2}$

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

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

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

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

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

Factor by the greatest common divisor $2$

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

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

$\frac{1}{2}\int\frac{1}{u-u^{3}}du$
5

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

$\frac{1}{2}\int\frac{1}{u-u^{3}}du$
6

Factor the polynomial $u-u^{3}$ by it's GCF: $u$

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

Calculate the power $\sqrt{1}$

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

Any expression multiplied by $1$ is equal to itself

$\frac{1}{2}\int\frac{1}{u\left(1+u\right)\left(1-u\right)}du$
7

Factor the difference of squares $\left(1-u^2\right)$ as the product of two conjugated binomials

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

Rewrite the fraction $\frac{1}{u\left(1+u\right)\left(1-u\right)}$ in $3$ simpler fractions using partial fraction decomposition

$\frac{1}{u\left(1+u\right)\left(1-u\right)}=\frac{A}{u}+\frac{B}{1+u}+\frac{C}{1-u}$
9

Find the values for the unknown coefficients: $A, B, C$. The first step is to multiply both sides of the equation from the previous step by $u\left(1+u\right)\left(1-u\right)$

$1=u\left(1+u\right)\left(1-u\right)\left(\frac{A}{u}+\frac{B}{1+u}+\frac{C}{1-u}\right)$
10

Multiplying polynomials

$1=\frac{uA\left(1+u\right)\left(1-u\right)}{u}+\frac{uB\left(1+u\right)\left(1-u\right)}{1+u}+\frac{uC\left(1+u\right)\left(1-u\right)}{1-u}$
11

Simplifying

$1=A\left(1+u\right)\left(1-u\right)+uB\left(1-u\right)+uC\left(1+u\right)$
12

Expand the polynomial

$1=A+Au-uA-u^2A+Bu-Bu^2+Cu+Cu^2$
13

Assigning values to $u$ we obtain the following system of equations

$\begin{matrix}1=A&\:\:\:\:\:\:\:(u=0) \\ 1=-2B&\:\:\:\:\:\:\:(u=-1) \\ 1=2C&\:\:\:\:\:\:\:(u=1)\end{matrix}$
14

Proceed to solve the system of linear equations

$\begin{matrix}1A & + & 0B & + & 0C & =1 \\ 0A & - & 2B & + & 0C & =1 \\ 0A & + & 0B & + & 2C & =1\end{matrix}$
15

Rewrite as a coefficient matrix

$\left(\begin{matrix}1 & 0 & 0 & 1 \\ 0 & -2 & 0 & 1 \\ 0 & 0 & 2 & 1\end{matrix}\right)$
16

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -\frac{1}{2} \\ 0 & 0 & 1 & \frac{1}{2}\end{matrix}\right)$
17

The integral of $\frac{1}{u\left(1+u\right)\left(1-u\right)}$ in decomposed fraction equals

$\frac{1}{2}\int\left(\frac{1}{u}+\frac{-\frac{1}{2}}{1+u}+\frac{\frac{1}{2}}{1-u}\right)du$
18

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

$\frac{1}{2}\int\frac{1}{u}du+\frac{1}{2}\int\frac{-\frac{1}{2}}{1+u}du+\frac{1}{2}\int\frac{\frac{1}{2}}{1-u}du$
19

We can solve the integral $\int\frac{-\frac{1}{2}}{1+u}du$ 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 $v$), which when substituted makes the integral easier. We see that $1+u$ it's a good candidate for substitution. Let's define a variable $v$ and assign it to the choosen part

$v=1+u$

Differentiate both sides of the equation $v=1+u$

$dv=\frac{d}{du}\left(1+u\right)$

Find the derivative

$\frac{d}{du}\left(1+u\right)$

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

$\frac{d}{du}\left(1\right)+\frac{d}{du}\left(u\right)$

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

$\frac{d}{du}\left(u\right)$

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

$1$
20

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

$dv=du$
21

Substituting $v$ and $du$ in the integral and simplify

$\frac{1}{2}\int\frac{1}{u}du+\frac{1}{2}\int\frac{-\frac{1}{2}}{v}dv+\frac{1}{2}\int\frac{\frac{1}{2}}{1-u}du$

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\frac{1}{2}\ln\left(u\right)$
22

The integral $\frac{1}{2}\int\frac{1}{u}du$ results in: $\frac{1}{2}\ln\left(u\right)$

$\frac{1}{2}\ln\left(u\right)$

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$-\frac{1}{4}\ln\left(v\right)$

Replace $v$ with the value that we assigned to it in the beginning: $1+u$

$-\frac{1}{4}\ln\left(1+u\right)$

Replace $u$ with the value that we assigned to it in the beginning: $\sin\left(x\right)$

$-\frac{1}{4}\ln\left(1+\sin\left(x\right)\right)$
23

The integral $\frac{1}{2}\int\frac{-\frac{1}{2}}{v}dv$ results in: $-\frac{1}{4}\ln\left(1+\sin\left(x\right)\right)$

$-\frac{1}{4}\ln\left(1+\sin\left(x\right)\right)$

Apply the formula: $\int\frac{n}{ax+b}dx$$=\frac{n}{a}\ln\left(ax+b\right)$, where $a=-1$, $b=1$, $x=u$ and $n=\frac{1}{2}$

$-\frac{1}{4}\ln\left(-u+1\right)$
24

The integral $\frac{1}{2}\int\frac{\frac{1}{2}}{1-u}du$ results in: $-\frac{1}{4}\ln\left(-u+1\right)$

$-\frac{1}{4}\ln\left(-u+1\right)$
25

Gather the results of all integrals

$\frac{1}{2}\ln\left(u\right)-\frac{1}{4}\ln\left(1+\sin\left(x\right)\right)-\frac{1}{4}\ln\left(-u+1\right)$

$\frac{1}{2}\ln\left(\sin\left(x\right)\right)-\frac{1}{4}\ln\left(1+\sin\left(x\right)\right)-\frac{1}{4}\ln\left(-\sin\left(x\right)+1\right)$
26

Replace $u$ with the value that we assigned to it in the beginning: $\sin\left(x\right)$

$\frac{1}{2}\ln\left(\sin\left(x\right)\right)-\frac{1}{4}\ln\left(1+\sin\left(x\right)\right)-\frac{1}{4}\ln\left(-\sin\left(x\right)+1\right)$
27

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

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

Final Answer

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