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

Step-by-step Solution

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Final Answer

$-\frac{1}{2}\ln\left(1-\tan\left(\frac{x}{2}\right)^{2}\right)+\frac{1}{2}\ln\left(\tan\left(\frac{x}{2}\right)\right)+C_0$
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Step-by-step Solution

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1

We can solve the integral $\int\frac{1}{2\sin\left(x\right)\cos\left(x\right)}dx$ by applying the method Weierstrass substitution (also known as tangent half-angle substitution) which converts an integral of trigonometric functions into a rational function of $t$ by setting the substitution

$t=\tan\left(\frac{x}{2}\right)$
2

Hence

$\sin x=\frac{2t}{1+t^{2}},\:\cos x=\frac{1-t^{2}}{1+t^{2}},\:\mathrm{and}\:\:dx=\frac{2}{1+t^{2}}dt$
3

Substituting in the original integral we get

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

Multiplying fractions $\frac{2t}{1+t^{2}} \times \frac{1-t^{2}}{1+t^{2}}$

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

Multiply $2$ times $2$

$\int\frac{1}{\frac{4t\left(1-t^{2}\right)}{\left(1+t^{2}\right)\cdot \left(1+t^{2}\right)}}\frac{2}{1+t^{2}}dt$

Divide fractions $\frac{2}{\frac{4\left(1-t^{2}\right)t}{1+t^{2}}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

$\int\frac{2\left(1+t^{2}\right)}{4\left(1-t^{2}\right)t}dt$

Cancel the fraction's common factor $2$

$\int\frac{1+t^{2}}{2\left(1-t^{2}\right)t}dt$
4

Simplifying

$\int\frac{1+t^{2}}{2\left(1-t^{2}\right)t}dt$
5

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

$\frac{1}{2}\int\frac{1+t^{2}}{\left(1-t^{2}\right)t}dt$
6

Rewrite the fraction $\frac{1+t^{2}}{\left(1-t^{2}\right)t}$ in $2$ simpler fractions using partial fraction decomposition

$\frac{1+t^{2}}{\left(1-t^{2}\right)t}=\frac{At+B}{1-t^{2}}+\frac{C}{t}$
7

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

$1+t^{2}=\left(1-t^{2}\right)t\left(\frac{At+B}{1-t^{2}}+\frac{C}{t}\right)$
8

Multiply both sides of the equality by $1$ to simplify the fractions

$1\left(1+t^{2}\right)=1\left(1-t^{2}\right)t\left(\frac{At+B}{1-t^{2}}+\frac{C}{t}\right)$
9

Multiplying polynomials

$1\left(1+t^{2}\right)=\frac{1\left(1-t^{2}\right)t\left(At+B\right)}{1-t^{2}}+\frac{1\left(1-t^{2}\right)tC}{t}$
10

Simplifying

$1\left(1+t^{2}\right)=t\left(At+B\right)+\left(1-t^{2}\right)C$
11

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

$\begin{matrix}2=A-B&\:\:\:\:\:\:\:(t=-1) \\ 2=A+B&\:\:\:\:\:\:\:(t=1) \\ 5=4A+2B-3C&\:\:\:\:\:\:\:(t=2)\end{matrix}$
12

Proceed to solve the system of linear equations

$\begin{matrix}1A & - & 1B & + & 0C & =2 \\ 1A & + & 1B & + & 0C & =2 \\ 4A & + & 2B & - & 3C & =5\end{matrix}$
13

Rewrite as a coefficient matrix

$\left(\begin{matrix}1 & -1 & 0 & 2 \\ 1 & 1 & 0 & 2 \\ 4 & 2 & -3 & 5\end{matrix}\right)$
14

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1\end{matrix}\right)$
15

The integral of $\frac{1+t^{2}}{\left(1-t^{2}\right)t}$ in decomposed fraction equals

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

Expand the integral $\int\left(\frac{2t}{1-t^{2}}+\frac{1}{t}\right)dt$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

$\frac{1}{2}\int\frac{2t}{1-t^{2}}dt+\frac{1}{2}\int\frac{1}{t}dt$

Taking the constant ($2$) out of the integral

$1\int\frac{t}{1-t^{2}}dt+\frac{1}{2}\int\frac{1}{t}dt$

Any expression multiplied by $1$ is equal to itself

$\int\frac{t}{1-t^{2}}dt+\frac{1}{2}\int\frac{1}{t}dt$
16

Simplify the expression inside the integral

$\int\frac{t}{1-t^{2}}dt+\frac{1}{2}\int\frac{1}{t}dt$
17

We can solve the integral $\int\frac{t}{1-t^{2}}dt$ 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 $1-t^{2}$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=1-t^{2}$

Differentiate both sides of the equation $u=1-t^{2}$

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

Find the derivative

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

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

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

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

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

The derivative of a function multiplied by a constant ($-1$) is equal to the constant times the derivative of the function

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

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

$-2t$
18

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

$du=-2tdt$
19

Isolate $dt$ in the previous equation

$\frac{du}{-2t}=dt$

Simplify the fraction $\frac{\frac{t}{u}}{-2t}$ by $t$

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

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

$-\frac{1}{2}\int\frac{1}{u}du$
20

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

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

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

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

Replace $u$ with the value that we assigned to it in the beginning: $1-t^{2}$

$-\frac{1}{2}\ln\left(1-t^{2}\right)$
21

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

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

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

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

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

$\frac{1}{2}\ln\left(t\right)$
23

Gather the results of all integrals

$-\frac{1}{2}\ln\left(1-t^{2}\right)+\frac{1}{2}\ln\left(t\right)$
24

Replace $t$ with the value that we assigned to it in the beginning: $\tan\left(\frac{x}{2}\right)$

$-\frac{1}{2}\ln\left(1-\tan\left(\frac{x}{2}\right)^{2}\right)+\frac{1}{2}\ln\left(\tan\left(\frac{x}{2}\right)\right)$
25

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

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

Final Answer

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

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 (1/2sinx)dx using basic integralsSolve integral of (1/2sinx)dx using u-substitutionSolve integral of (1/2sinx)dx using integration by parts

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|◻|
θ
=
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<
>=
<=
sin
cos
tan
cot
sec
csc

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acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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