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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acosh
atanh
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Final answer to the problem

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

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  • Integrate by partial fractions
  • Integrate by substitution
  • Integrate by parts
  • Integrate using tabular integration
  • Integrate by trigonometric substitution
  • Weierstrass Substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
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Simplify $2\sin\left(x\right)\cos\left(x\right)$ using the trigonometric identity: $\sin(2x)=2\sin(x)\cos(x)$

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

Simplify the fraction $\frac{2\sin\left(2x\right)}{2}$ by $2$

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

Applying the cosecant identity: $\displaystyle\csc\left(\theta\right)=\frac{1}{\sin\left(\theta\right)}$

$\csc\left(2x\right)$
1

Simplify $\frac{1}{2\sin\left(x\right)\cos\left(x\right)}$ into $\csc\left(2x\right)$ by applying trigonometric identities

$\int\csc\left(2x\right)dx$
2

We can solve the integral $\int\csc\left(2x\right)dx$ 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 $1$

$2$
3

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

$du=2dx$

Rearrange the equation

$2dx=du$

Divide both sides of the equation by $2$

$dx=\frac{du}{2}$
4

Isolate $dx$ in the previous equation

$dx=\frac{du}{2}$
5

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

$\int\frac{\csc\left(u\right)}{2}du$
6

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

$\frac{1}{2}\int\csc\left(u\right)du$
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The integral of $\csc(x)$ is $-\ln(\csc(x)+\cot(x))$

$-\left(\frac{1}{2}\right)\ln\left|\csc\left(u\right)+\cot\left(u\right)\right|$
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Multiply the fraction and term in $-\left(\frac{1}{2}\right)\ln\left|\csc\left(u\right)+\cot\left(u\right)\right|$

$-\frac{1}{2}\ln\left|\csc\left(u\right)+\cot\left(u\right)\right|$

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

$-\frac{1}{2}\ln\left|\csc\left(2x\right)+\cot\left(2x\right)\right|$
9

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

$-\frac{1}{2}\ln\left|\csc\left(2x\right)+\cot\left(2x\right)\right|$
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Simplify $\csc\left(2x\right)+\cot\left(2x\right)$ using trigonometric identities

$-\frac{1}{2}\ln\left|\cot\left(x\right)\right|$
11

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

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

Final answer to the problem

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

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Function Plot

Plotting: $-\frac{1}{2}\ln\left(\cot\left(x\right)\right)+C_0$

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5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
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

How to improve your answer:

Main Topic: Trigonometric Integrals

Integrals that contain trigonometric functions and their powers.

Used Formulas

See formulas (2)

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