## anonymous one year ago Using one computer algebra system, it was found that ∫dx/ (1+sin x ) = (sin x - 1)/ cos x and using another computer system, this integral is equal to 2 sin (x/2) / cos(x/2) + sin(x/2)... How do I reconcile these two answers?

1. Loser66

You meant: $$\dfrac {sinx -1}{cosx}$$ vs $$\dfrac{sin(x/2)}{cos(x/2)+sin(x/2)}$$ right?

2. anonymous

$$\color{slate}{\displaystyle \int \frac{1}{\sin(x)+1}dx}$$ $$\color{slate}{\displaystyle \int \frac{1(\sin x - 1)}{(\sin x+1)(\sin x - 1)}dx}$$ $$\color{slate}{\displaystyle \int \frac{\sin x - 1}{\sin^2x-1}dx}$$ $$\color{slate}{\displaystyle -\int \frac{\sin x - 1}{\cos^2x}dx}$$ $$\color{slate}{\displaystyle -\int \left(\frac{\sin x }{\cos^2x}-\frac{\ 1}{\cos^2x}\right)dx}$$ $$\color{slate}{\displaystyle -\int \left(\tan x \sec x -\sec^2x\right)dx}$$ $$\color{slate}{\displaystyle \int \left(\sec^2x-\tan x \sec x \right)dx}$$

3. anonymous

I hope this is a helpful approach.

4. anonymous

yeah, sec^2x = tan^2x+1 i forget what derivative is what, but whether you need sec^2x or tan^2x, you got it.

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