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

## Step-by-step solution

Problem to solve:

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

Solving method

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

Divide $1$ by $2$

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

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

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

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

Multiplying the fraction by $\sec\left(x\right)$

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

Applying the trigonometric identity: $\displaystyle\sec\left(\theta\right)=\frac{1}{\cos\left(\theta\right)}$

$\frac{1}{2}\int\frac{\sec\left(x\right)}{\sin\left(x\right)}dx$
3

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

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

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

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

Find the derivative

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

The derivative of the tangent of a function is equal to secant squared of that function times the derivative of that function, in other words, if ${f(x) = tan(x)}$, then ${f'(x) = sec^2(x)\cdot D_x(x)}$

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

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

$\sec\left(x\right)^2$
4

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

$du=\sec\left(x\right)^2dx$
5

Isolate $dx$ in the previous equation

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

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

$\int\frac{\sec\left(x\right)}{\sin\left(x\right)\sec\left(x\right)^2}du$

Simplify the fraction by $\sec\left(x\right)$

$\int\frac{1}{\sin\left(x\right)\sec\left(x\right)}du$

Applying the trigonometric identity: $\displaystyle\frac{1}{\sec(\theta)}=\cos(\theta)$

$\int\frac{\cos\left(x\right)}{\sin\left(x\right)}du$

Apply the trigonometric identity: $\frac{\cos\left(x\right)}{\sin\left(x\right)}$$=\cot\left(x\right)$

$\int\cot\left(x\right)du$

As the original substitution was $u=\tan\left(x\right)$, we can substitute the inverse of $tan$, which is $cot$, for $\frac{1}{u}$

$\int\frac{1}{u}du$
6

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

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

$\frac{1}{2}\cdot 1\ln\left|u\right|$

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

$\frac{1}{2}\ln\left|u\right|$
7

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

$\frac{1}{2}\ln\left|u\right|$

$\frac{1}{2}\ln\left|\tan\left(x\right)\right|$
8

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

$\frac{1}{2}\ln\left|\tan\left(x\right)\right|$
9

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

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

$\frac{1}{2}\ln\left|\tan\left(x\right)\right|+C_0$
$\int\frac{1}{2sin\left(x\right)cos\left(x\right)}dx$