Solving Integral: ∫ 1/(1 + tan x) dx

❓Solving Integral: ∫ 1/(1 + tan x) dx
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[@justabunga1]

This is an improper integral since x=π/2 is undefined at the point. Therefore, the upper limit of integration must changed in terms of limits. There is another method that you can look at is to let u=tan(x) so du=(sec(x))^2dx. Notice the identity (tan(x))^2+1=(sec(x))^2. By the u sub, it turns the integral as the integral from 0 to infinity of 1/((1+u)(1+u^2))du. Put this as a partial fraction decomposition by setting 1/((1+u)(1+u^2))=A/(1+u)+(Bu+C)/(1+u^2). Solving for A, B, and C algebraically gives us A=1/2, B=1/2, and C=-1/2. Take out 1/2 outside of the integral as 1/2(integral from 0 to infinity of 1/(u+1)+(-u+1)/(1+u^2)du). Separate the second term as -u/(1+u^2)+1/(1+u^2). Integrating that term gives us -1/2ln(1+u^2)+arctan(u). We know the integral of 1/(u+1)du is ln(|u+1|). Multiply everything by 1/2 for all of terms gives 1/2ln(|u+1|)-1/4ln(1+u^2)+1/2arctan(u). Using properties of logarithms in the first and second term should be ln((|u+1|)^(1/2)/(1+u^2)^(1/4))+1/2arctan(u). Evaluating this in terms of limits from 0 to infinity should be (ln(1)+(1/2)(π/2))-(0+0)=(0+π/4)-0=π/4.
We can also look at it another way by changing the integrand in terms of sin(x) and cos(x) without changing the limits of integration. The integrand should be cos(x)/(cos(x)+sin(x)). The tricky method is to make this integrand the same. If I add (cos(x)+sin(x))/(cos(x)+sin(x)), this equals to 1. Integrating this gives us x. We know that the derivative of sin(x)+cos(x) is cos(x)-sin(x). Adding (cos(x)-sin(x))/(cos(x)+sin(x)) to the term (cos(x)+sin(x))/(cos(x)+sin(x))=2cos(x)/(cos(x)+sin(x)), so the integral of (cos(x)-sin(x))/(cos(x)+sin(x))dx=ln(|cos(x)+sin(x)|). Because there is a coefficient 2 in front of the cos(x), you can divide everything by 2 to get the integrand the same, which the whole integral should be (x+ln(|cos(x)+sin(x)|)/2+C. Evaluating this from 0 to π/2 is π/4 (ignoring the +C since you are evaluating the definite integral).

[@user-kl9ch4pk5t]

You can solve it without any integration. The function is like a triangle with height 1 and base pi/2 ( got it from putting the limit ) . Then area of triangle is 1/2×pi/2×1 = Pi/4

[@mohammadsenanali]

That's great. Unique observation.

[@BadrBoutara]

in the second 0,39 i didnt understand how did 2(1+tan(x)) came

[@mohammadsenanali]

Consider I = int (1/(1 + tan x))dx and I = int (tan x/(1 + tan x))dx. Adding both, we get 2I = int (1 + tan x)/(1 + tan x)dx. Dividing by 2 on both sides, we find I = (1/2) * int (1 + tan x)dx, leading to the presence of 2(1 + tan x). Cheers!

[@BadrBoutara]

@@mohammadsenanali wonderful

[@waqarpunisher8992]

​@@mohammadsenanaliin what world is 1/(1+tanx)=tanx/(1+tanx)?

[@mohammadsenanali]

@@waqarpunisher8992 The world in which the domain is considered between 0 to pi/2