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

Step-by-step Solution

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Final answer to the problem

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

How should I solve this problem?

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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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1

We can solve the integral $\int\frac{1}{\sin\left(x\right)-1}dx$ 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

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

Combine $\frac{2t}{1+t^{2}}-1$ in a single fraction

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

Multiplying fractions $\frac{1}{\frac{2t-\left(1+t^{2}\right)}{1+t^{2}}} \times \frac{2}{1+t^{2}}$

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

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

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

Simplify the fraction $\frac{2\left(1+t^{2}\right)}{\left(2t-\left(1+t^{2}\right)\right)\left(1+t^{2}\right)}$ by $1+t^{2}$

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

Simplifying

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

Solve the product $-\left(1+t^{2}\right)$

$\frac{2}{2t-1-t^{2}}$

The trinomial $2t+1+1t^{2}$ is a perfect square trinomial, because it's discriminant is equal to zero

$\Delta=b^2-4ac=2^2-4\left(-1\right)\left(-1\right) = 0$

Using the perfect square trinomial formula

$a^2+2ab+b^2=(a+b)^2,\:where\:a=\sqrt{1t^{2}}\:and\:b=\sqrt{1}$

Factoring the perfect square trinomial

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

Rewrite the expression $\frac{2}{2t-\left(1+t^{2}\right)}$ inside the integral in factored form

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

Simplify the division $2$ by $-1$

$\int\frac{-2}{\left(t-1\right)^{2}}dt$
7

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

$u=t-1$

Differentiate both sides of the equation $u=t-1$

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

Find the derivative

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

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

$1$
8

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

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

$\int\frac{-2}{u^{2}}du$
10

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\int-2u^{-2}du$
11

The integral of a function times a constant ($-2$) is equal to the constant times the integral of the function

$-2\int u^{-2}du$
12

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $-2$

$-2\left(\frac{u^{-1}}{-1}\right)$
13

Simplify the fraction $-2\left(\frac{u^{-1}}{-1}\right)$

$2u^{-1}$

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

$2\left(t-1\right)^{-1}$
14

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

$2\left(t-1\right)^{-1}$
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Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

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

$\frac{2}{\tan\left(\frac{x}{2}\right)-1}$
17

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

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

Final answer to the problem

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

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Function Plot

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

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a
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m
n
u
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x
y
z
.
(◻)
+
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×
◻/◻
/
÷
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

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