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

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The integral of \( \frac{1}{1+\tan x} \) can be solved by substituting \( u = \tan x \), leading to \( dx = \frac{1}{1+u^2} du \). This transforms the integral into a form suitable for partial fraction decomposition. The resulting expression is simplified to \( \frac{1}{1+u} \cdot \frac{1}{1+u^2} \), which can be decomposed into manageable parts. The final solution is expressed as \( \frac{1}{2} \ln(1+\tan x) - \frac{1}{4} \ln(1+\tan^2 x) + \frac{1}{2} x + C \). This method effectively demonstrates the application of substitution and partial fraction decomposition in solving the integral.
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\( \int\frac{dx}{1+tanx} \)
 
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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 'Proving that convexity implies second order derivative being positive'

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