MHB How to solve the integral of 1+tanx using partial fraction decomposition?

  • Thread starter Thread starter Lorena_Santoro
  • Start date Start date
  • Tags Tags
    Approach
Click For Summary
SUMMARY

The integral of \( \frac{1}{1+\tan x} \) can be solved using substitution and partial fraction decomposition. By substituting \( u=\tan x \), the integral transforms into \( \int \frac{1}{1+u} \cdot \frac{1}{1+u^2} \, du \). The solution yields \( \frac{1}{2}\ln(1+\tan x) - \frac{1}{4}\ln(1+\tan^2 x) + \frac{1}{2}x + C \), where \( C \) is the constant of integration. This method effectively combines logarithmic and arctangent functions to arrive at the final result.

PREREQUISITES
  • Understanding of integral calculus
  • Familiarity with substitution methods in integration
  • Knowledge of partial fraction decomposition
  • Basic proficiency in trigonometric functions and their properties
NEXT STEPS
  • Study advanced techniques in integral calculus
  • Explore more examples of partial fraction decomposition
  • Learn about the properties of logarithmic and arctangent functions
  • Investigate the applications of integrals in real-world problems
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on calculus, as well as educators seeking effective methods for teaching integration techniques.

Lorena_Santoro
Messages
22
Reaction score
0
\( \int\frac{dx}{1+tanx} \)
 
Physics news on Phys.org
Substitute $u=\tan x$, which means $x=\arctan u$ and $dx=\frac{1}{1+u^2}\,du$.
Follow up with partial fraction decomposition.
$$\begin{align}\int\frac{dx}{1+\tan x}&=\int \frac{1}{1+u}\cdot\frac{1}{1+u^2}\,du
=\int\Big(\frac{\frac 12}{1+u}+\frac{-\frac 12 u +\frac 12}{1+u^2}\Big)\,du
=\frac 12\ln(1+u)-\frac 14\ln(1+u^2)+\frac 12\arctan u + C \\
&=\frac 12\ln(1+\tan x)-\frac 14\ln(1+\tan^2 x) +\frac 12 x + C\end{align}$$
 
Klaas van Aarsen said:
Substitute $u=\tan x$, which means $x=\arctan u$ and $dx=\frac{1}{1+u^2}\,du$.
Follow up with partial fraction decomposition.
$$\begin{align}\int\frac{dx}{1+\tan x}&=\int \frac{1}{1+u}\cdot\frac{1}{1+u^2}\,du
=\int\Big(\frac{\frac 12}{1+u}+\frac{-\frac 12 u +\frac 12}{1+u^2}\Big)\,du
=\frac 12\ln(1+u)-\frac 14\ln(1+u^2)+\frac 12\arctan u + C \\
&=\frac 12\ln(1+\tan x)-\frac 14\ln(1+\tan^2 x) +\frac 12 x + C\end{align}$$
Thank you very much!
 

Thread 'How does this derivative work? And why the speed of light remains?'

Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...
View full post »

Similar threads

MHB 206.08.05.44 partial fraction decomposition
  • · Replies 6 ·
Replies
6
Views
2K
Integrals that keep me up at night
  • · Replies 22 ·
Replies
22
Views
3K
Graduate Multiplying divergent integrals using Hardy fields approach
  • · Replies 3 ·
Replies
3
Views
2K
High School Integration by parts of inverse sine, a solved exercise, some doubts...
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad Partial Fraction Decomposition With Quadratic Term
  • · Replies 3 ·
Replies
3
Views
2K
MHB Integrate 3/(x(2-sqrt(x))) - No Partial Fractions
  • · Replies 5 ·
Replies
5
Views
2K
A doubt in Partial fraction decomposition
  • · Replies 8 ·
Replies
8
Views
2K
MHB Is This Partial Fraction Decomposition Set Up Correct?
  • · Replies 8 ·
Replies
8
Views
2K
Undergrad Different Substitution for Solving an Integral
  • · Replies 6 ·
Replies
6
Views
2K
MHB Would Partial Fractions Simplify This Integral?
  • · Replies 8 ·
Replies
8
Views
2K