0

When trying to integrate $\frac{1}{1+sin(x)}$, I used two methods.

  1. Weierstrass substitution gives me
    $\frac{-2}{\tan(\frac{x}{2}) + 1}$

  2. Multiplying by the conjugate gives me
    $\tan x - \frac{1}{\cos x}$

Yet when I plug in numbers, I get different results

$\tan(0) - \frac{1}{\cos(0)} = -1-\frac{2}{t+\tan(\frac{0}{2})} = -2$

$\tan(90) - \frac{1}{\cos(90)} =$ undefined$ -\frac{2}{t+\tan(\frac{90}{2})} = -1$

https://www.youtube.com/watch?v=CKuE9hFDec8 https://www.youtube.com/watch?v=6pGLK7iPpmA

These two videos by the same person did the same two methods and got the same answers, so I don't think I made any calculation errors.

Piita
  • 965
user12456500
  • 1
  • $$\lim_{x\to 90^o-}\left(\tan x-\frac1{\cos x}\right)$$ might exist, even if the value is undefined when $x=90^o.$ – Thomas Andrews May 28 '23 at 20:15
  • One thing is sure : $$\tan 90°-\frac1{\cos 90°}= \infty - \infty$$ a case where you surely cannot conclude that it is equal to $-1$ ! – Jean Marie May 28 '23 at 20:26

1 Answers1

1

You can check that

$$\frac{-2}{\tan(x/2)+1}=\tan x-\frac{1}{\cos x}\color{red}{-1}$$

So as far as indefinite integration goes, they are considered equivalent.

K.defaoite
  • 12,536