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We tried to do an integral with the universal trigonometric substitution $$\int \frac{1}{(1+\sin x)}\, \mathrm dx$$
Meaning, we substituted: $ t = \tan \frac{x}{2} \Rightarrow$
$$\int \frac{1}{(1+\sin x)}\, \mathrm dx = \int \frac{\frac{2}{1+t^2}}{1+\frac{2t}{1+t^2}}\, \mathrm dt = \int \frac{2}{(1+t)^2}\, \mathrm dt = \frac{-2}{1+t} = \frac{-2}{1+\tan \frac{x}{2}} + C$$
But the answer is: $$ \tan x - \frac{1}{\cos x} + C $$
What did we do wrong?
$$\begin{align} \frac{-2}{1+\tan \frac{x}{2}} &= \frac{-2\cos \frac{x}{2}}{\cos\frac{x}{2} + \sin \frac{x}{2}}\\ &= \frac{-2\cos \frac{x}{2}(\cos \frac{x}{2}-\sin\frac{x}{2})}{(\cos \frac{x}{2}+\sin\frac{x}{2})(\cos \frac{x}{2}-\sin\frac{x}{2})}\\ &= \frac{\sin x - 2 \cos^2 \frac{x}{2}}{\cos x}\\ &= \tan x - \frac{\cos x + 1}{\cos x}\\ &= \tan x - \frac{1}{\cos x} - 1 \end{align}$$
You did nothing wrong, you just got another representation of the same family of functions.
An alternative method of integration first involves multiplying the numerator and denominator by $1-\sin x$: this gives $$\begin{align*} \int \frac{1}{1+\sin x} \, dx &= \int \frac{1 - \sin x}{1 - \sin^2 x} \, dx \\ &= \int \frac{1 - \sin x}{\cos^2 x} \, dx \\ &= \int \sec^2 x - \frac{\sin x}{\cos^2 x} \, dx \\ &= \tan x - \int \frac{-du}{u^2}, \quad u = \cos x, du = -\sin x \, dx \\ &= \tan x - \frac{1}{u} + C \\ &= \tan x - \sec x + C. \end{align*}$$ To see the equivalence of this form with the expression $$-\frac{2}{1+\tan \frac{x}{2}} + C,$$ consider their difference, with $\theta = x/2$: $$\begin{align*} \tan x - \sec x + \frac{2}{1 + \tan \frac{x}{2}} &= \frac{\sin 2\theta - 1}{\cos 2\theta} + \frac{2}{1 + \tan \theta} \\ &= \frac{\sin 2\theta - 1}{\cos 2\theta} + \frac{2 \cos \theta}{\sin \theta + \cos \theta} \\ &= \frac{\sin 2\theta - 1 + 2 \cos\theta(\cos \theta - \sin \theta)}{\cos 2\theta} \\ &= \frac{2 \cos^2 \theta - 1}{\cos 2\theta} \\ &= 1. \end{align*}$$ Thus their difference is constant, and both are antiderivatives.
Another approach would be to first multiply the integrand by $\dfrac{1-\sin x}{1-\sin x}$ to get $$\begin{aligned}\int \frac{1}{1+\sin x}\,dx &= \int\frac{1-\sin x}{1-\sin^2 x}\,dx\\ &= \int\frac{1-\sin x}{\cos^2x}\,dx \\ &= \int\frac{1}{\cos^2 x}-\frac{\sin x}{\cos^2 x}\,dx \\ &= \int \sec^2x -\frac{\sin x}{\cos^2x}\,dx\\ &= \ldots\end{aligned}$$ The first term is trivial to integrate, and the second integral requires the substitution $u=\cos x$. I leave it to you to finish things off.
Another answer is $$ \int\frac{1}{1+\sin x} dx=-\frac{\cos x}{1+\sin x}+c $$ To show this we write $$ \tan x-\sec x=\frac{\sin x-1}{\cos x}= \frac{(\sin x-1)\cos x}{\cos^2 x}= \frac{(\sin x-1)\cos x}{1-\sin^2 x}= \frac{(\sin x-1)\cos x}{(1-\sin x)(1+\sin x)}= -\frac{\cos x}{1+\sin x}+c $$
Another answer:
Let $x=\dfrac\pi2-2u$. Then $dx=-2du$ and so the integral is $$\int\frac{-2du}{1+\cos(2u)}=-\int \sec^2(u)du=-\tan u+c=\cdots$$ where $c$ is constant.
We used $\cos 2u=2\cos^2 u-1$ and $\sec x=\cos^{-1}x$.
