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Exercise

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

Step-by-step Solution

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

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

$\frac{2t}{1-t^{2}}+\frac{1}{t}$
7

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

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$u=1-t^{2}$
9

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 finding the derivative of the equation above

$du=-2tdt$
10

Isolate $dt$ in the previous equation

$\frac{du}{-2t}=dt$
11

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

$-\frac{1}{2}\int\frac{1}{u}du+\frac{1}{2}\int\frac{1}{t}dt$
12

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

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

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

$\frac{1}{2}\ln\left(t\right)$
14

Gather the results of all integrals

$-\frac{1}{2}\ln\left|1-t^{2}\right|+\frac{1}{2}\ln\left|t\right|$
15

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 exercise

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

Try other ways to solve this exercise

  • 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
  • Load more...
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Main Topic: Weierstrass Substitution

In integral calculus, the Weierstrass substitution or tangent half angle substitution is a method for solving integrals, which converts a rational expression of trigonometric functions into an algebraic rational function, which can be easier to integrate. The Weierstrass substitution is very useful for integrals that involve a simple rational expression with sine and/or cosine in the denominator.

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