Evaluate the integral: I = integral dx/(1 + sin(x))

Evaluate the integral: $$ I = \int \frac{dx}{1 + \sin(x)}. $$

Step 1 · Use the Weierstrass substitution

To solve this integral, we use the substitution:

$$ t = \tan\left(\frac{x}{2}\right). $$

This substitution is a common method when dealing with integrals involving $\sin(x)$ and $\cos(x)$. The following identities hold:

$$ dx = \frac{2}{1+t^2}\,dt, \qquad \sin(x) = \frac{2t}{1+t^2}. $$

Step 2 · Simplify $1+\sin x$

Substituting the expression for $\sin(x)$:

$$ 1+\sin(x) = 1 + \frac{2t}{1+t^2} = \frac{1+t^2+2t}{1+t^2} = \frac{(1+t)^2}{1+t^2}. $$

This is the key simplification: the denominator becomes a perfect square divided by $1+t^2$.

Step 3 · Substitute into the integral

Substituting $dx = \dfrac{2}{1+t^2}\,dt$ and $1+\sin(x)=\dfrac{(1+t)^2}{1+t^2}$ into the integral:

$$ I = \int \frac{\frac{2}{1+t^2}}{\frac{(1+t)^2}{1+t^2}}\,dt = \int \frac{2}{(1+t)^2}\,dt. $$

The $1+t^2$ factors cancel exactly, leaving a simple power of $(1+t)$.

Step 4 · Integrate

Using the power rule for integration:

$$ \int \frac{2}{(1+t)^2}\,dt = 2\int (1+t)^{-2}\,dt = 2\cdot \left(-\frac{1}{1+t}\right) = -\frac{2}{1+t}. $$

Thus:

$$ I = -\frac{2}{1+t} + C. $$

Step 5 · Substitute back $t = \tan(x/2)$

Finally, substituting back $t = \tan\left(\dfrac{x}{2}\right)$, we obtain the result:

$$ I = -\frac{2}{1+\tan\left(\frac{x}{2}\right)} + C. $$

Summary Table

Additional Notes

• The domain of $I$ excludes $x=-\tfrac\pi2+2k\pi$ for integer $k$, where $1+\sin x=0$, and points where $\tan(x/2)$ itself is undefined ($x=\pi+2k\pi$).
• An equivalent closed form is $I=\tan x-\sec x+C$, obtained by multiplying the original integrand by $\frac{1-\sin x}{1-\sin x}$ and using $1-\sin^2x=\cos^2x$; the two forms differ only by an additive constant.
• The result can be checked by differentiating $-\dfrac{2}{1+\tan(x/2)}$ with the chain rule and confirming it returns $\dfrac{1}{1+\sin x}$.