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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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##  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
Can't find a method? Tell us so we can add it.
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

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

Simplifying

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

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

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

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

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

$\begin{matrix}2=A-B&\:\:\:\:\:\:\:(t=-1) \\ 2=A+B&\:\:\:\:\:\:\:(t=1) \\ 5=4A+2B-3C&\:\:\:\:\:\:\:(t=2)\end{matrix}$
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Proceed to solve the system of linear equations

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

$\left(\begin{matrix}1 & -1 & 0 & 2 \\ 1 & 1 & 0 & 2 \\ 4 & 2 & -3 & 5\end{matrix}\right)$
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Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1\end{matrix}\right)$
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The integral of $\frac{1+t^{2}}{\left(1-t^{2}\right)t}$ in decomposed fraction equals

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

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

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

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

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

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

##  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

SnapXam A2

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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

###  Main Topic: Trigonometric Integrals

Integrals that contain trigonometric functions and their powers.