How to integrate $\int\frac{1}{1+\sin(x)} dx$

Question

$$\int\frac{1}{1+\sin(x)} dx$$

The integration techniques I know are:

Inspection, power rule, integral of basic functions (also trig), and substitution

But none of them help me solve this.

How can I solve this integral with the techniques I know so far?

Answer 1

Hint: Note that $$\frac{1}{1+\sin x}=\frac{1-\sin x}{1-\sin^2 x}=\frac{1-\sin x}{\cos ^2x}=\sec^2x-\sec x\tan x.$$ Then recall that $$(\tan x)'=\sec^2x,\quad (\sec x)'=\sec x\tan x.$$

Answer 2

HINT:

$$1+\sin x=1+\cos\left(\dfrac\pi2-x\right)=2\cos^2\left(\dfrac\pi4-\dfrac x2\right)$$

Answer 3

HINT

Let $t=\tan \frac{x}{2}$. Now using Weierstrass substitution

Note that the integral becomes $$\int \frac{1}{1+\sin x} \mathrm{d}x=\int \frac{2}{t^2+2t+1} \mathrm {d} t$$

Can you take it from here?

Answer 4

$$\displaylines{ \sin x = {{2\tan x/2} \over {1 + {{\tan }^2}x/2}} \cr 1 + {\tan ^2}x/2 = {\sec ^2}x/2 \cr}$$ use these formulas to convert the integral to required form.

Answer 5

$$\frac 1{1+\sin x}=\frac{1}{(\sin \frac x2+\cos\frac x2)^2}=\frac{1}{\sin^2\frac x2(1+\cot\frac x2)^2}=\frac{1+\cot^2\frac x2}{(1+\cot\frac x2)^2}$$ set $u=1+\cot\frac x2$, thus $$I=2\int-\frac{1}{u^2}du=\frac2u+c$$

Answer 6

Use Identity
$$\displaylines{ \sin 2x = {{2\tan x} \over {1 + {{\tan }^2}x}} \cr 1 + {\tan ^2}x = {\sec ^2}x \cr} $$
Given integral
$$\displaylines{ I= \int {{1 \over {1 + \sin x}}dx} = \int {{1 \over {1 + {{2\tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}}}} \cr = \int {{{1 + {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2} + 2\tan {x \over 2}}}} dx \cr} $$
Now take
$$\displaylines{ \tan {x \over 2} = t \cr {1 \over 2}{\sec ^2}{x \over 2}dx = dt \cr {\sec ^2}{x \over 2}dx = 2dt \cr} $$ this will modify the integral as, $$\displaylines{ I = \int {{{2dt} \over {1 + {t^2} + 2t}}} \cr = 2\int {{{dt} \over {1 + {t^2} + 2t}}} \cr} $$
You can integrate this integral. Work on it. Good Luck!

Answer 7

You'll probably want to do it by substitution.

Let the denominator equal to say $u$.

You then have a simple integral of the form $\dfrac1u$ which resolves to $\ln u + c$.

Answer 8

\begin{align} I=\int\frac{dx}{1+\sin x}&=\int d\left(\frac{-\cos x}{1+\sin x}\right)=\frac{-\cos x}{1+\sin x}+C \end{align}