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

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Solution

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Final answer to the problem

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

How should I solve this problem?

  • Weierstrass Substitution
  • Integrate by partial fractions
  • Integrate by substitution
  • Integrate by parts
  • Integrate using tabular integration
  • Integrate by trigonometric substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
  • FOIL Method
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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

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

$\sin x=\frac{2t}{1+t^{2}},\:\cos x=\frac{1-t^{2}}{1+t^{2}},\:\mathrm{and}\:\:dx=\frac{2}{1+t^{2}}dt$
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Substituting in the original integral we get

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

$\int\frac{1+t^{2}}{2\left(1-t^{2}\right)t}dt$
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Take the constant $\frac{1}{2}$ out of the integral

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

$\frac{2t}{1-t^{2}}+\frac{1}{t}$
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Simplify the expression

$\int\frac{t}{1-t^{2}}dt+\frac{1}{2}\int\frac{1}{t}dt$
8

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

$u=1-t^{2}$
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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

$du=-2tdt$
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Isolate $dt$ in the previous equation

$\frac{du}{-2t}=dt$
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Substituting $u$ and $dt$ in the integral and simplify

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

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

$\frac{1}{2}\ln\left(t\right)$
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Gather the results of all integrals

$-\frac{1}{2}\ln\left|1-t^{2}\right|+\frac{1}{2}\ln\left|t\right|$
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Replace $t$ with the value that we assigned to it in the beginning: $\tan\left(\frac{x}{2}\right)$

$-\frac{1}{2}\ln\left|1-\tan\left(\frac{x}{2}\right)^{2}\right|+\frac{1}{2}\ln\left|\tan\left(\frac{x}{2}\right)\right|$
16

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 to the problem

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

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Function Plot

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

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x
y
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.
(◻)
+
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◻/◻
/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
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

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Main Topic: Trigonometric Integrals

Integrals that contain trigonometric functions and their powers.

Used Formulas

See formulas (2)

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