Integrate $\int \frac {dx}{1+\sin x}$

Question

Integrate $\int \dfrac {dx}{1+\sin x}$

My attempt: $$\begin{align}&=\int \dfrac {dx}{1+\sin x}\\ &=\int{\dfrac{dx}{1+\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1+\tan ^2\left( \dfrac{x}{2} \right)}}}\\ &=\int{\dfrac{1+\tan ^2\left( \dfrac{x}{2} \right)}{\left( 1+\tan \left( \dfrac{x}{2} \right) \right) ^2}}\,dx\end{align}$$

Answer 1

Hint:$$\int\frac1{1+\sin x}\,\mathrm dx=\int\frac{1-\sin x}{\cos^2x}\,\mathrm dx=\int\frac1{\cos^2x}\,\mathrm dx-\int\frac{\sin x}{\cos^2x}\,\mathrm dx.$$

Answer 2

Hint:

$$\dfrac1{1+\sin x}=\dfrac1{1+\cos\left(\dfrac\pi2-x\right)}=\dfrac{\sec^2\left(\dfrac\pi4-\dfrac x2\right)}2$$

Alternatively, $$\dfrac1{1+\sin x}=\dfrac{\sec^2\dfrac x2}{\left(1+\tan\dfrac x2\right)^2}$$

Answer 3

See Tangent_half-angle_substitution

If $z = \tan \big(\frac x2 \big)$, then $dx = \dfrac{2 dz}{1+z^2}$ and $\sin(x) = \dfrac{2z}{1+z^2}$

\begin{align} \int\frac{dx}{1+\sin x} &= \int\frac{1}{1+\sin x} \cdot dx \\ &= \int \dfrac{1}{\left(1+\dfrac{2z}{(1+z^2)}\right)} \cdot \dfrac{2 dz}{1+z^2} \\ &= \int \dfrac{2 \, dz}{(1+z)^2} \\ &= -\dfrac{2}{1+z} + C \\ &= -\dfrac{2}{1 + \tan(\frac x2)} \\ &= \dfrac{-2}{1 + \dfrac{\sin x}{1+\cos x}} \\ &= -2\dfrac{1+\cos x}{1 + \cos x + \sin x} \end{align}

Answer 4

$$\int\frac{1}{1+sin(x)}dx$$ Apply u-substitution:$u=tan(\frac x2)$

$$=\int \frac{2}{u^2+1+2u}$$ $$=2\int \frac{1}{u^2+1+2u}$$ $$=2\int\frac{1}{(u+1)^2}$$ Again apply u-substitution:$v=u+1$ $$=2\int\frac{1}{v^2}dv$$ $$=2\int\ v^{-2}dv$$ $$=2\frac{v^{-2+1}}{-2+1}$$ Substitute back $v=u+1,u=tan\frac {x}{2}$

$$=-\frac{2}{tan\frac x2+1}$$ $$\int\frac{1}{1+sin(x)}dx=-\frac{2}{tan\frac x2+1}+C$$

Answer 5

I am continuing from where you left -

$$\int \dfrac {dx}{1+\sin x}$$ $$=\int \dfrac {dx}{1+\dfrac {2\tan (\dfrac {x}{2})}{1+\tan^2 (\dfrac {x}{2})}}$$ $$=\int \dfrac {1+\tan^2 (\dfrac {x}{2})}{(1+\tan (\dfrac {x}{2}))^2}\,dx$$ $$=\int \dfrac {1+(\dfrac {\sin^2 (\dfrac {x}{2})}{\cos^2 (\dfrac {x}{2})})}{(1+\tan (\dfrac {x}{2}))^2}\,dx$$ $$=\int \dfrac {\dfrac{\sin^2(\dfrac{x}{2})+\cos^2(\dfrac{x}{2})}{\cos^2(\dfrac{x}{2})}}{(1+\tan (\dfrac {x}{2}))^2}\,dx$$ $$=\int \dfrac {\dfrac{1}{\cos^2(\dfrac{x}{2})}}{(1+\tan (\dfrac {x}{2}))^2}\,dx$$ $$=\int \dfrac {\sec^2(\dfrac{x}{2})}{(1+\tan (\dfrac {x}{2}))^2}\,dx$$ $$ Let,$$ $$\quad \quad \quad \quad \quad \quad \quad \quad 1+\tan(\dfrac{x}{2})=z$$ $$\quad \quad \quad \quad \quad \quad \quad \quad \implies(\sec^2(\dfrac{x}{2}))dx=2dz$$ $$=\int 2\dfrac {dz}{z^2}$$ $$=\dfrac{-2}{z}+c$$ $$=-\dfrac {2}{(1+\tan (\dfrac {x}{2}))}+c$$